MANISH KUMAR
PHYSICS
GRAVITATION INTRODUCTION: Sir Isaac Newton observed that the earth, basides attracting mall objects on or near its surface, attracts planet and other heavenly bodies far away from. It. He further assumed that the force of attracting exists between a objects big or small, not only on the surface of the Earth, but anywhere and everywhere in the universe irrespective of the distances separating them. He named this force of attracting between any two bodies as the force of graviton
GRAVITATION Example: Consider a girl whirling a stone along a circular path. If the girl releases the stone at some point, the stone flies of along the tangent, at that point on the circular path. Let us discuss this observation carefully. Before the release of thread, the stone was moving with a certain uniform speed and changed its direction at every point. Because of the change in direction, it moved with a variable velocity and has some definite acceleration. The force that causes this acceleration and makes the stone move along the circular path, acts towards the centre, i.e., towards the hand of the girl. This force is called centripetal force. When the thread is released, the stone does not experience the centripetal force and files off along a straight line. This straight line is always tangent to the circular path.
NEWTON’S UNIVERSAL LAW OF GRAVITATION Every particle in the universe every other particle with of force which is directly proportional to the product of two masses and inversely proportional to the square of the distance between them. The direction of the force is along the line joining two masses. If m1 and m2 are the masses of two bodies separated by distance d and F is the force of attraction between them, then F m1m2 1 F d2 m1m2 d2 G m1m2 or F Where G is a constant of proportionally and known as the of universal gravitation and d2 is equal to
F
MANISH KUMAR
PHYSICS
Fd2 = G m1m2 Fd 2 G= m1m2 If m1 = m2 = 1kg and d = 1m, then
F x 12 or G = F 1x1 i.e. Universal gravitation constant is the force of attraction (in newton) between two bodies of mass 1 (kg) each lying 1 (m) distance apart. G=
UNIT OF GRAVITATION CONSTANT G=
Fd 2 m1m 2
In SI unit G =
Nm 2 Nm 2 = = Nm2 kg-2 2 kg x kg kg
In CGS unit of dyn cm-2 g-2 The value of G = 6.67 x 10-11 N m2 kg-2 or 6.67 x 10-8 dyn cm2 g-2 The value of G was found by henry cavendish.
IMPROTANCE OF UNIVERSAL LWO OF GRAVITATION The Universal low of gravitation given by Newton has explained successfully several phenomena. For 1.
example The gravitation force attraction of the Earth is responsible for binding all terrestrial object on the Earth
2. 3.
The gravitation force of the Earth is responsible for holding the atmosphere aground the Earth. The gravitation force of the Earth is also responsible for holding the atmosphere around the Earth.
4. 5.
The flow of water in rivers is also due to gravitational force the Earth on water. The moon revolves around the Earth on account of gravitational pull of the Earth on the Moon. Even all
6.
artificial satellites All the planets revolve around the Sun due to gravitational pull of the sun on the planets . Thus, gravitation force alone is responsible for holding our solar system.
7. 8.
The tides formed by the rising and falling of water level in the ocean are due to the gravitational force attraction, which the Sun and the Moon exert on sea water. The predictions about solar and lunar eclipses made on the basis of this law always come out to be true.
GRAVITATIONAL FROCE BETWEEN LIGHT OBJECT AND HEAVY OBJECTS The formula applied for calculating gravitational force between light objects and heavy object is the.
MANISH KUMAR
PHYSICS
G m1m2 .Let us take three cases: r2 1. When two of mass 1 kg each are 1 metre apart . Sol. i.e. m1 = m1 = 1 kg, r = 1 m Taking G = 6.67 x 10-11 Nm2/kg2, we obtain gravitational force of attraction
same, i.e.F =
G m1m2 6.67 x 10 11 x 1x 1 F= = r2 (1) 2 Which is extremely small. Hence we conclude that though every pair of two object exert gravitation pull o each other, yet they cannot move towards eachother because this gravitational pull is to weak. 2. When a body of mass 1 kg is held on the surface of Earth. Sol. Here, m1 = 1 kg m2 = mass of Earth = 6 x 1024 kg r = distance of body from centre of Earth = radius of Earth = 6400 km = 6.4 x 103 km = 6.4 x 106 m Gravitational force of attraction between the body and Earth. F=
G m1m2 6.67 x 10 11 x 1 x 6 x 10 24 = = 9.8 N r2 (6.4 x 10 6 ) 2
It means that the Earth exerts a gravitational force of 9.8 N on a body mass one kg. force is much larger compared to the force when both the bodies are lighter. That is why when a body dropped from a height it falls to the Earth. 3. When both the bodie are heavy Sol. Let us calculate gravitational force of attraction between Earth and the Moon. Mass of Earth, m1 = 6 x 1024 kg Mass of Moon, m2 = 7.4 x 1022 kg Distance between Earth and Moon, r = 3.84 x 105 kg = 3.84 x 108 Gravitational constant, G = 6.67 x 10-11Nm2 /kg2 The gravitational force between Earth and Moon. F=
G m1m2 6.67 x 10 11 x 6 x 10 24 x 7.4 x 10 22 = 2.01 x10 20 N 8 2 2 r (3.84 x 10 )
Which is really. It is this large gravitational force exerted by Earth on Moon, which the moon revolve around the Earth. Ex. Let us find force of attraction between block lying 1m apart. Let the mass of each block is 40 kg. Sol. F=? m1 = 40 kg
MANISH KUMAR
PHYSICS
m2 = 40 kg d =1m G = 6.67 x 10-11 N m2 kg-2 F=
G m1m2 6.67 x 10 11 x 1 x 6 x 10 24 = = 9.8 N r2 (6.4 x 10 6 ) 2
G m1m2 6.67 x 10 11 x 40 x 40 = 1x1 d2 -2 F = 1.0672 x 10 N
F=
Or
KEPLER’S LEWS OF PLANETRY MOTION
1.
2.
Johannes kepler was a 16th astronomer who established three laws which govern the motion of planets (around the sun). These are kwon as kepler’s lows of planetary motion. The same laws also describe the motion of satellites (like the moon) around the planets (like the earth). The Kaplar’s laws of planetary motion are give below. KEPLER’S FIRST LAW : The planets move in elliptical around the sun, with at one of the to foci of the elliptical orbit.
KEPLER’S SECOND LOW : Each revolves around the sun in such a way that the line joining plant to the sun sweeps over equal areas in equal intervals of time.
MANISH KUMAR 3.
PHYSICS
KEPLER’S THIRD LAW : The cube of the mean distance of a plant the sun is directly proportional the square of time it takes to move around the sun. The law can be expressed as : r3 T2 Or r3 = constant x T2 r3 = constant T2 Where r = Mean distance of planet from the sun and T = Time period of the planet (around the sun) Through Kepler gave the laws of planetary motion but he could not give a theory to explain the motion of planets. It was Newton who showed that the cause of the motion of planets is the gravitation force which the sun exerts on them. In fact, Newton used the Kapler’s third law of planetary motion of develop the law of universal gravitation
Or
MANISH KUMAR
PHYSICS
NEWTON’S INVERSE-SQUARE RULE The force between two bodies is inversely proportional the square of distance between them is called the inverse-square rule. 1 r2
F
Consider planet of mass m moving with a velocity (of speed) v around the sun in circular robit of radius r , centripetal force F acts on the orbiting planet (due to the sun) which is given by :
mv 2 r
F=
The mass m of a given planet is constant
v2 F r If the planet takes time T to complete one revolution (of 2r) around the sun, then its velocity v is given by: 2ππ T
v=
The factor 2 is a constant r T
v=
Now, taking square on both sides
r2 v = 2 T 2
If we multiply as well as divide the right side of this relation by r v3 = The factor
r3 is constant by Keplar’s third law. T3
v2 by putting
r3 1 x T3 r
1 r
1 v2 in place of v2 in relation F r r
MANISH KUMAR F
PHYSICS
1 1 or F 2 rxr r
NEWTON’S THIRD LAWOF MOTION AND GRAVITATION The Newton third of motion also good for the force of gravitation. This means that when earth exerts a force of attraction on an object, then the object also exerts an equal force on the earth, in the opposite direction. According to Newton’s second law, orce = Mass x Acceleration F = me Force Acceleration = Mass F Or a= m The mass of earth is very large and acceleration produced in the earth very very small and cannot detected with even the most accurate instrument available to us.
FREE FALL The falling of a body (or object) from a height towards the earth under the gravitational force of earth (with on other force acting on it) is called free fall
ACCELERATION DUE TO GRAVITY When a body dropped from a certain height . it falls with a constant acceleration. This uniform acceleration produced in a freely falling body due to the gravitational pull of the earth’s is known as acceleration due to gravity and it is denoted by g. “the acceleration of a body due to attraction of earth its centre is calle acceleration due to gravity The value of g = 9.8 m/s2
DETERMINATION OF VALUE OF g When a body of mass m is dropped from a certain distance R from centre of earth mass M, then the exerted by the earth on the body is GMm F= …… (i) R2 Let this force produced an acceleration a in mass in F = ma From (i) and (ii), GMm Ma = R
MANISH KUMAR
PHYSICS
GM R2 For bodies falling near the surface of earth, this acceleration is called acceleration due to gravity and is represented by g. GM g= 2 R
Or
a=
MANISH KUMAR
PHYSICS
Where M is the mass of the earth i.e. 6 x 1024 kg and R, radius the earth i.e. 6.4 x 106 m GM 6.67 x 10 11 (6 x 10 24 ) = R2 (6.4 x 10 24 )
g=
Or
g = 9.8 ms-2 or nearly 10 ms-2
VALUE OF g ON MOON Mass of moon = 7.4 x 1022 kg and its radius = = 1,740 km Or
R = 1,740,000 m = 1,74 x 106 m g=
GM 6.67 x 10 11 (7.4 x 10 22 ) = = 1.63 ms-2 2 6 2 R (1.74 x 10 )
We have already seen that acceleration due to gravity does not depend upon mass of falling body. Mass of the earth We can determine mass of the earth from equation (i) g=
GM R2
or M =
gR 2 G
Mass of the earth M =
9.8(6.4 x 106 ) 6.66 x 1011
Or M = 5.99 x 1024 kg
AVERAGE DENSITY OF THE EARTH It can also be determined from equation (i) above 4 G πR 3 GM 4 3 g= = 2 = G Rd 2 R R 3 or
d=
3g G4R
Taking the earth to be a sphere of radius R
3 x 9.8 6.66 x 10 x 4 x (6.4x 106 ) 2
d=
Or
d = 5.5 x 103 kg m-3
-11
Calculation of acceleration due to gravity on the moon and to prove that it is 1/6th of the acceleration due to gravity on the earth. Mass of the moon (M) = 7-4 x 1022 kg
MANISH KUMAR Radius of the moon (R) = 1.74 x 106 m
PHYSICS
MANISH KUMAR
PHYSICS
Gravitational constant (G) = 6.7 x 10-11 Nm2/kg2 Acceleration due to gravity on the moon, g = gMoon =
6.7 x 10-11 Nm 2 kg -2 x 7.4 x 10 22 (1.7 x 106 m) 2
gMoon =
1011 22 6.7 x 7.4 x N/kg 1.74 x 1.74 1012
GM R2
gMoon = 1.63 ms-2
g Moon g Earth
gMoon =
=
1.63ms 2 1 = approx 9.81ms 2 6
1 gEath 6
VERIATION OF ACCELERATION DUE TO GRAVITY ON THE SURFACE OF THE EARTH The acceleration due to gravity of the earth at the poles is 9.83 m/s2 and at the equator is 9.78 m/s2. The generally accepted value of the acceleration due tc gravity, that is 9.81 m/s2 is the average acceleration. It is not the same at all places of the Earth. Furthermore, the average acceleration due to gravity is, with respect to the sea level. Its value can further vary with the height or depth. In general, the acceleration due to gravity at sea level maximum at the geographic poles and minimum at equator. The value of acceleration due to gravity decrease when one moves : (i) from the poles towards equator, (ii) away from the earth as on hills, balloons, spaceships, etc., (iii) into the earth say in deep mines.
MEMORISE 1.
The acceleration due to gravity of a planet depend on its mass and its radius. Its value is high if mass is large and radius is small.
2.
The value of g at the surface of is 9.8 ms-2 on an avarge.
3.
The value of decreases with height.
4.
The value of g decreases with depth.
gh =
GM (R h)2
MANISH KUMAR 5. 6.
The value of g is more at poles and less at equator. The value of zero at the centre of the earth.
PHYSICS
MANISH KUMAR
PHYSICS
Ex. Mass of 1kg falling towards it with acceleration of 9.8 ms-2. The force acting on it will be Sol. F = ma = 1 x 9.8 = 9.8 N Let A be the acceleration with which earth rises towards the object, then F = MA Where M = mass of the earth M = 6 x 1024 kg F A= M 9.8 Or A= = 1.63 x 10-24 ms-2 6 x 1024 Ex. Calculate the force of gravitation due child of mass 25 kg on his fat mother mass 75 kg if the istance between their centres is 1m from each other. Given G = (20/3) x 10-11 Nm2 kg-2 20 Sol. Here m1 = 25 kg ; m2 d = 1 m ; G = x 10-11 Nm2 Kg-2 3 Gm1m 2 Using F = d2 Or
Q. A
F
=
20 x 1011 x 75 x 75 3 x (1) 2
Or F = 12,500 x 10-11 Or F = 1.25 x 10-7 N mass of 45 kg is attracted by mass 15 kg lying at a distance of 2 m with a force 1.67 x 10-8 N. Find the alue of G.
GRVITATION AND GRAVITY Gravitational is the force of attraction between any two bodies whereas gravity is the force of attraction between two bodies when one of the two bodies earth. Hence gravity is special of gravitation.
EQUATION OF MOATION FREELY FALLING BODIES When the bodies are falling under influence of gravity, they experience acceleration g.i.e. 9.8 ms-2. However when these are going up against gravity, they move with retardation of 9.8 m-2 All the equation of motion already read by us are valid for freely falling body with the difference that a replaced by g. For motion vertically upwards (a) is replased by (-g). The equation of motion v = u + at Replace a = g v = u + gt When body falls in downward v = u – gt When body through upward
MANISH KUMAR s = ut +
PHYSICS 1 2 at 2
MANISH KUMAR Relace
PHYSICS
a=g& s=h H = ut +
1 2 gt 2
v2 – u2 = 2as Relace
s=h v2 – u2 = 2gh
Ex.1
A body drops a stone from the edge of the root. If passes a window 2m high in 0.1s. How far is the root above the top of the window ?
Sol.
Let a stone be dropped from the edge of the roof A. Let it passes over B with a velocity say u. Consider motion BC. u = ? ; a = 9.8 ms-2 ; s = h = 2m ; t = 0.1s Using s = ut +
1 2 gt , we have 2
2 = u (0.1) +
1 x 9.8 (0.1) 2 2
2 = 0.1u + 0.049 0.1u = u – 2 – 0.049 or
u=
1.915 1951ms 1 0.1
Root is 19.4 m above the window. Ex2.
A ball thrown up is caught by the ghrower after 4s. whit what velocity was it thrown up ? How high did go ? Where it after 3 s? (g = 9.8 s-2)
Sol.
Since the time of going up is the same that of commg down, therefore time of going = 4/2 = 2s. starts upward with velocity u. Hereu = ? ; a = 9.8 m s-2 ; t = 2s; v = 0 (at the top); s = h Using
v = u + at
Or
u = u -9.8 x 2
Or
u = 19.6 m s-1
Again
v2 – u2 = 2as 0(19.6)2 = 2 (-9.8) h h = 19.6 m
MANISH KUMAR
Ex.
Sol.
PHYSICS
After 2s. it starts coming downwards (starting with u = 0) Considering motion. u = 0 ; a = 9.8 m s-2 ; t = 3 – 2 = 1s ; s = ? 1 s = ut + at2 2 1 or s = 0 + x 9.8 (1)2 4.9 m from top. 2 Coconut is hanging on a tree at a height of 15 m from the ground A body launches a projectile ertically upwards with a velocity of 20 m s-1 . After what time the projecting pass by coconut ? Explain the two answers in this problem. Here u = 20 m s-1; a = - 10 m ; s = 15 m ; t = ? 1 Using s = ut + at2, we have 2 1 15 = 20t + (-10) t2 2 Dividing throughout by 5, we have 3 = 4t – t2 or t2 = 4t + 3 = 0 or (t – 1) (t - 3) = 0 t – 1 = 0 or t = 1s or t – 3 = 0 or t = 3s After 1s, it will cross coconut while going up and after 3 s while coming down.
MASS The amount of matter contained in a body is called its mass Or The measure of the quantity of matter in a body is called its mass. The mass of a body is a scalar quantity. It is independent of surrounding and the position of the body. It is a constant quality for a given body. Mass is measured in kilograms (kg) in SI system.
WEIGHT Everybody on the surface of earth is attracted towards the centre of earth. The force of attraction depends upon the mass of the body and acceleration due to gravity. The weight of the body is the force with which it is attracted towards the centre of the earth. We know F = ma The acceleration produced by the froe of attraction of the earth is known as acceleration due to gravity .e.g. F = ma = mg But by definition this force is equal to the weight of the body i.e. F = W.
MANISH KUMAR W = mg SI unit weight is Newton (N) and CGS, it is measured in dyne (dyn).
PHYSICS
MANISH KUMAR
PHYSICS
PRACTICAL UNITS In SI, the weight is also measured in kg f of kg wt. Therefore, kilogram froe or kilogram weight is force with which a mass of 1 kg is attracted by centre of earth 1 kg f = 1 kg wt = 9.8 N In CGS, the practical unit of weight is grams force or a wt org g f or 1 g wt is force with which a mass 1 g of attracted b the centre of the earth. 1 g f = 1 g wt = 980 dyn WEIGHT OF A BODY ON THE SURFACE OF EARTH AT DIFFERENT PLACE Since the weight of the body depends upon mass and acceleration due to gravity, g. The value of g change from place to place and h ; pce the weight of the body is different at different place. EFFECT OF SHAPE OF THE EARTH ON WEIGHT The earth is not perfectly spherical. Its radius at poles is less thi GM
THRUST AND PRESSURE THRUST : Force acting normally on a surface is called the thrust. Thrust is a vector quantity and is measured in the unit of force i.e. newton (N) PRESSURE : The thrust acting on unit area of the surface is called the pressure If a thrust F or acts on a area A. then pressure (P) =
Thrust(F) Area( A )
F A A sharp knife cuts easily than a blunt knife by applying the same force.
P=
Ex. Ex.
A sharp needle pressed against our skin pierces it. But a blunt object with a contact area does not affect the skin when pressed against it with the same force. UNITS OF PRESSURE. The SI unit of pressure is called passed (Pa) in honour of Blaise Pascal. 1 Pa = 1 N/m2 One pascal is defined is the pressure exerted on a surface of area of 1 m2 by a thrust of 1 N (acting normally no it) Other unit of pressure are bat and millibar (m bar) where 1 ber = 105 N/m2 and 1 milliber = 102 N/m2 It is a common practice in meteorology t measure atmospheric in bars and millibars. Further, 1 atmospheric pressure (1a atm) = 101.3 k pa = 1.013 ber = 1013 m ber Fig. (a) shows how to calculate pressure exerted by a brick of mass 3 kg : (a) when standing on end ; (b) when lying flat. The total force or thrust exerted is the same in both the cases.
MANISH KUMAR
PHYSICS
In Fig. (a), A1 = 5 cm x 10 cm = 50 cm2 = 50 x 10-4 m2 and F1 = 3 KG wt = 3 kg x 10 m/s2 = 30N
Thus, Pressure exerted, P1 =
F1 A1
=
30 N 6 x 103 N/m2 = 6 x 103 pa 4 2 50 x 10 m
In Fig (b), A2 = 10 cm x 20 cm = 200 cm2 200 x 10-4 m2 and F2 = 3 kg wt = 3 kg x 10 m/s2 = 200 x 10-4 and Thus, pressure exerted, P2 =
F2 30N = = 1.5 x 103 N/m2 = 1.5 x 103 pa 4 2 A 2 200 x10 m
SUME INTERESTING ASPECTS OF PRESSURE 1.
The foundation of a building or a dam as large surface area so that the pressure exerted by it on the ground is less. This done to prevent the sinking of the building or the into the ground
2.
The tyres of as bus or a truck have large width than those of a car. Further. The number of tyres of heavy vehlcles is more than four. This is done to enable the tyres to carry more weight and to prevent sinking into ground.
3.
Nails and pins have pointed ends so that these can be fixed with minimum force because the pressure on the pointed ends would be large.
4.
A sleeping mattress is so designed that when you lie on it,a large area of your body comes in its contact. This reduces the pressure on the body and sleeping become comfortable.
5.
Wide wooden sleepers are kept below railway lines to reduce pressure on the railway and prevent them zrom sinking into ground.
MANISH KUMAR
PHYSICS
DENSITY Density of a substance is defined as its mass per unit volume. Unit of density. Since mass (M) is measured in kilogram (kg) and the volume (v) is measured in metre3 (m3), the different for different substance as give in Table.
Material (a) Solids : 1. Aluminium 2. Cast iron 3. Copper 4. Gold 5. Lead 6. Silver 7. Tin 8. Tungsten 9. Uranium 10 Zinc (b) Liquids: 1. Crude oil 2. Ethyl alcohol 3. Gasoline 4. Heavy water 5. Mercury 6. Niroglycerine 7. Sea water 8. Sulp huric acid 9. Toluene 10. Water (c) Gases : 1. air 2. ammonia 3. Argon 4. Carbon dioxide 5. Chlorine 6. Oxygen 7. Ozone
Density (kg/m3) 2.7 x 103 7.0 x 103 8.93 x 103 19.31 x 103 11.35 x 103 10.5 x 103 7.29 x 103 19.34 x 103 19.1 x 103 7.15 x 103 0.76 – 0.85 x 103 0.79 x 103 1.26 x 103 1.1086 x 103 13-55 x 103 1.6 x 103 1.01 -1.03 x 103 1.83 x 103 0.866 x 103 0.99823 x 103 1.293 0.771 1.783 1.977 3.22 1.429 2.139
RELATIVE DENSTIY In many cases, instead of dealing with the density of a substance, it is preferable to consider the number time the substance is as dense as water. This is called the relative density Relative density of a substance is defined as the ratio of its density to that of water at 4o C.
MANISH KUMAR Thus, Denstiy of substacne Denstiy of waterat 4 o C Linit of relative density Since relative density is a ratio of two similar quantities, it no unit. Further, Denstiy of substacne relative density = Denstiy of waterat 4 o C
Relative density =
PHYSICS
MANISH KUMAR
PHYSICS
mass of subs tan ce / volume of subs tan ce mass of / volume of water at 4 0 C If the volume of a given substance is equal to the volume of water at 40C, mass of subs tan ce relative density mass of an equal volume of water at 4 0 C Relative density can also be defined as the radio between the mass of the substance and the mass of an eque volume of waer at 4oC
PRESSURE IN FLUIEDS A substance which can flow is called a fluid. All liquids and gases thus fluids. We know that a solid exerts pressure on a surface due to its weight. Similarly, a fluid exerts on the container in which it constaineo due to weight. However, unlike a solid, a fluid exerts pressure in all direction. A fluid contained in a vessel exerts pressure at all points vessel in all direction. All the streams of water reach almost the same distance in the air.
PASCAL’S LAW
In an enclosed fluid, if pressure is changed in any part of the fluid, then this change of pressure is ransmitted undiminished to all the other parts of the fluid.
BUOYANCY When a body is partially of wholly immerged in a liquid, an upward force on it which is called upthrust or buoyant force, The propertly of liquid responsible for this force is called buoyancy.
MANISH KUMAR
PHYSICS
ARCIMEDES PRINCIPLE
(a)
(b)
(c)
Consider a container C1 filled with water upto the level from where pipe P extends out. The other end of pipe P opens to small container C2 placed on a weighing balance which measure 00.00 [ after the placement of the container C2 ]. A block B hangs on a spring balance S which shows a reading of 7 kg. If we partially immerse the block in water we observe some water flows out from C1 to C2 through P. The weight of machine shows reading 1 kg and the loss of reading in sprig balance is 7 – 6 = 1kg. This means weight of water displaced by the block is equal to loss in weight of block. Now we completely immerse the block in water, we observe that the weight of water displaced by the block is kg and the reading in spring balance is 2 kg. The loss of weight of block is 7- 2 = 5 kg. Again we reach same conclusion that weight of water displaced by the block is equal to the loss in weight of block What happen when the block is further immersed? No more water will be displayed by the block and therefore reading shows by weighing machine and spring balance remains unchanged. Why the spring balance shows a loss in weight of the block when. Fig. A block of 7 kg hanging on a the block immersed in water ? This is because of buoyant forcespring balance acting vertically upwards. The loss in weight is equal to the buoyant force. zz
CONCLUSION 1 : Buoyant force volume of liquid displaced (V). If two bodies of different material have same volume buoyant force acting on them, when completely immersed in water, is same. Instead of water if we take a liquid light than Mass We know that Density Volume Mass = Density x volume For lighter liquid, the mass of the liquid displaced is less even when the volume displaced is the same.
CONCLUSION 2 : Buoyant force Density of liquid
MANISH KUMAR i.e., Buoyant force d It has also been found that buoyant force also depend on the acceleration due to gravity
PHYSICS
MANISH KUMAR CONCLUSION 3 :
PHYSICS
Buoyant force g If we combine all the tree, we gat Buoyant force dg Buoyant force mg
[mass = V xd]
Buoyant force Weight of the liquid displaced Note : A body placed in a gaseous medium is also by the upthrust equal to the weight of the gas displaced.
The a bove facts has been summarised in Archimedes principle which states that the upward force acting on a solid body which is partially or completely immersed in a fluid, is equal to the weight of the fluid displaced. This upward force is called buoyant force or upthrust.
APPLICATIONS OF ARCHIMEDES PRINCIPLE 1. In designing ships and submarines. 2.
Lactometer is based on the Archimedes principle. It is used to determine the purity of sample of milk.
3.
Hydrometer or also based on the Archimedes. It is used to determine the density of liquid
MANISH KUMAR
PHYSICS
SOLVED EXEMPLE 1. What force of gravitation between two point masses of 1 kg and 2 kg kept 1 m apart ? Sol. m = 1 kg, m2 = 2 kg r = 1m m1m 2 1x 2 = 6.67 x 10-11 x 2 = 13.34 x 10-11 N. 2 r 1 This is an extremely small force.
F=G
2.
Calculate the force of gravitation between the earth and sun. Mass of Earth = 6 x 1024 kg, Mass of sun = 2 x 1030 kg. The distance between the two is 1.5 x 1011m.
Sol. F =
Gm1m 2 r2
6.67 6 2 = 1.5 1.5 3.
10 24 10 30 11 = 35.57 x 1021 N. 11 10 10
Write down the expression for acceleration experienced by a particle on the surface of the moon due to gravitational force on the moon. Find the ratio of this acceleration to that experienced by same particle on the surface of the earth. If the acceleration due to gravity on the earth is 9.8 ms-2, What is acceleration particle on the moon’s surface ? mass of moon = 7.3 x 1022 kg; Mass of Earth = 6 x 1024 kg. Radius of moon 1.74 x 106 m. Radius of earth = 64 x 106 m.
Sol. Acceleration on moon gm =
GM m Rm2
Also acceleration on the surface of Earth
gE
GM E R2E
Dividing (1) by (2)
gm GMm RE2 M = x = 2m 2 ge GMm REMm RE =
7.3 x 10 22 x 6.4 x 10 6 x 6.4 x 10 6 1.74 x 10 6 x 1.74 x 10 6 x 6 x10 24
=
7.3 x 6.4 x 10 6 x 6.4 x 10 6 1.74 x 1.74 x 6 x 100
MANISH KUMAR
PHYSICS
gm = 0.16 gE
gm = 0.16 x gE = 0.16x 9.8 = 1.57 ms-2 4.
Find the value of acceleration due to gravity a height of (a) 6400 km, 12,800 km from the surface of the earth. Radius of earth is 6400 km.
Sol. We know that gh = Also g =
GM (R h)2
GM R2
gh R2 = g (R h)2
R2 gh = g 2 (R h)
2
(a) g = 9.8 ms-2, R = 6400 km, R + h = 6400 + 6400 = 12800 km 2
2
9.8 6400 1 (gh)A 9.8 = 9.8 = = 2.45 ms-2 4 12800 2 2
6400 (b) (gh)B = 9.8 = 9.8 6400 12800
5.
2
1 3 = 1.09
As h increases gh decreases. This is because gh and h are inversely related.
A particle is thrown up vertically with a velocity of 50 m/s. (a) What will be its velocity at the highest pint of is journey ? (b) How high would the particle rise ? (c) What time would it take to reach the highest point.
Sol. At the highest point the velocity with a velocity be zero. Considering activity A to B Using V
= u + at
0 = 50 – 9.8 x t Also
t =
50 = 5.1 s 9.8
v2 – u2 = 2as 02 – (50)2 = 2 (-9.8) x s
MANISH KUMAR 5=
6.
PHYSICS
50 x 50 127.5 m 2 9.8
With reference to the above sample problem, (a) find the time particle takes from the highest point back to the initial point (b) Find the velocity with which the particle the initial point
Sol. The date is given in the adjacent figure. Considering activity B to A Using
v2 – u2 = 2as v2 - 02 = 5 (9.8) (127.5) ] v = 50 m/s
7.
Also
v = u + at
50 = 0 + 9.8 (t) ]
t = 5.1 s
A ball is dropped from the top of a tower 40 m high. What is its velocity when it has covered 20 m ? What would be its velocity when it hits the ground? Take g = 10m/s2.
Sol. Let the point B be at a height of 20 m. Activity from A toB. u1 = 0, a1 = 10 ms-2 s1= 20 m, v1 = ? v12 - u12 = 2a1s2
v12 -02 = 2 (10) (20)
v12 = 202 v1 = 20 m/s
Activity from A to C : C is a point on the ground u2 = 0. a2 = 10 ms-2. s2 = 40 m, v2 ?
8.
v22 - u22 = 2a1s1
v22 - 02 = 2 (10) (40)
v22 = 800
v2 = 28.28 ms-1
A body thrown up with a speed 29.4 ms-1 (a) What is its speed after (i) t = 1, (ii) t = 2 s and (iii) t = 3 s. (b) What is its height after (i) t = 1 s, (ii) t = 2 s and (iii) t = 3 s.
Sol. (a)
(i)
u = 29.4 ms-1, a = -9.8 ms-2, t1 = 1 s, v1 = ?
MANISH KUMAR
PHYSICS
v1 = u A + at1 = 29.4 + (- 9.8) X 1 19.6 ms-1 (ii)
u = 29.4 ms, a = -98 ms-2, t2 = 2 s, v2 = ? v2 = u + at2 = 29.4 + (-9.8) x 2 = 9.8 ms-1
MANISH KUMAR
PHYSICS
(iii) u = 29.4 m s-1, a = 9.8 ms-2, t3 = 3 s, v3 = ? v3 = u + at3 (b)
(i)
= 29.4 + (-9.8) x 3 = 0 u = 29.4 ms-1, a = -9.8 ms-2 s1 = h1, t1 = 1s
(ii)
1 2 1 at1 = 29.4 x 1 + (-9.8) x = 24.5 cm 2 2 u = 29.4 ms-1, a = -9.8 ms-2, s2 = h2, t2 = 2 s
h1= ut1+
1 2 1 at2 = 29.4 x 2 + (-9.8) x 22 = 39.2 m 2 2 -1 -2 (iii) u = 29. 4 ms a = -9.8 ms , s3 = h3, t3 = 3 s
h2 = ut2 +
1 3 1 at2 = 29.4 x 3 + (-9.8) x 32 = 44.1 m 2 2 What is weight of a person whose mass is 50 kg.
h3 = ut3 +
9.
Sol. The weight of the person W = mg = 50 x 9.8 = 490 N Note : The gravitational unit of force is kg- (kilogram force) or kg-we (kilogram weight) 1 kg-wt = 9.8 N = 1 kg – f 490 N = 50 kg-f 10. Weight of a girl is294 N. Find her mass. Sol. W = mg 294 = m x 9.8 294 = 30 kg 9.8 Weight of an object is 294 N on the surface of the earth. What is it’s a height of 200 km from the surface of
M=
11.
the earth. Radius of the earth = 6400 km. Wh = mgh 2 2 R R = mg gh g R h R h
Here mg = 294 N, R = 6400 km, h = 200 km 2
6400 wh = 294 = 294 6400 200
2
6400 6600 = 276.45 N
Note : Weight decreases with increase of height from the surface of the earth. 12.
The gravitational force change when (i) distance between them is reduced to half ?
MANISH KUMAR (ii) the mass of each object is quadruped ?
PHYSICS
MANISH KUMAR
PHYSICS
Sol. (i) According of Newton’s low gravitational, gravitation force F between two objects distance r apart is 1 F 2 r 1 When distance between them is reduced to half. i.e. r = r/2, then force, F’ 2 r 2 2 F' r r Thus, = 2= = 4 or F’ = 4 F F r (r / 2)2 i.e. force becomes 4 times its previous value. (ii) Again, according to Newton’s law of gravitation, the gravitational force between two object of mass m1 and m2 is F m1m2 when mass of each object is quadrupled, m’1 = 4m1 and m’2 = 4m2 The force, F’ m’1 m’2 F' m'1 m' 2 ( 4m1 )( 4m 2 ) Thus, = = 16 m1m 2 F m1m 2 Or F = 16 F i.e. force becomes 16 times is previous value 13. A sphere of mass 40 kg is attracted by a second sphere of mass 15 kg when their canters 320 cm apart, with of force of 0.1 milligram weight. Calculate the value of gravitational constant. Sol. Here m1 = 40 kg, m2 = 15 kg 0 From r = 20cm = = 2 x 10-1 m 100 F = 0.1 milligram weight = 0.1 x 10-3 gram weight = 10-4 x 10-3 kg wt = 10-7 x 9.8 N (1 kg wt = 9.8 N) Gm1m 2 From F = r2 F x r2 10 7 x 9.8 x (2 x 10 1 )2 G= = m1m 2 40 x 15 -11 2 G = 6.53 x 10 Nm / kg2 14. Calculate the force of force of gravity acting on your friend of mass 0 kg. Given of earth = 6 x 1024 kg and radius of Earth = 6.4 kg, M = 6 x 1024 kg Sol. Here, m = 60 kg, M = 6 x 1024 kg R = 6.4 x 106 m, F = ? G = 6.67 x 10-11 Nm2/kg2 GMm Thus F = R2
MANISH KUMAR or or
6.67 x10 11 x 6 x 10 24 x 60 6.4 x 10 6 F = 58.62 N F=
PHYSICS
MANISH KUMAR
PHYSICS
15.
A particle is thrown up vertically with a velocity of 50 m/s. What will e velocity highest point of the journey How high would the particle rise ? what time would it take to reach point ? Take = 10 m/s2 Sol. Here, initial velocity u = 50 m/s final velocity v=? height covered, h=? time taken. t =? At the highest point, final velocity v = 0 From v2 – u2 = 2 gh, 0 – (50)2 = 2 (-10) h where g = - 10 m/s2 upward journey. 2500 h= 125m 20 From v = u + gt. Or 0 = 50 + (-10) t t = 50/10 5s 16. A force of 15 N is uniformly distributed over area of 150 m2. Find the pressure pascals Sol. Here, force, F = 15 N area, A = 150 cm2 = 150 x 10-4 m2 (1 cm = 10-2 m, 1 cm2 = 10-4m2) F 15N Thus, pressure. P = = = 1000 pa A 150 x 10 4 m2 17. How much force should be applied on an area of 1 cm2 to gat pressure of 15 Pa ? Sol. Here. area. A = 1 cm2 = 10-4 m2 Pressure, P = 15 pa = 15 N/m2 F As P= , F = P x A = (15 N/m2) x (10-4 m2) 15 x 10-3 N A 18. A block weight 1.0 kg is in the shape of length 10 cm. it kept on a horizontal table. Find the pressure on the portion of the table where the block is kept. Sol. Here, force acting on the table, F = 1.0 kg = 10 N Area of the table on which is this force acts, A = 10 cm x 10 cm = 100 cm2 = 100 x 10-4 m2 = 10-2 m2 (1 cm2 = 10-4 m2) F 10N Pressure on the table, P = = = 1000 pa A 10 2 m 2 19. The pressure due to atmosphere is 1.013 x 105 pa. Find the force exerted by the atmosphere on the top surface of a table 2.0 m long and 1.0 m wide Sol. Here, pressure due to the atmosphere, P = 1.013 x 105 Pa = 1.013 x 105 N/m2 area on which atmospheric pressure acts, A = 2.0 m x 1.0 m = 2.0 m2 Thus, force exerted by the atmosphere, F = PA = (1.013 x 105 N/m2) x (2.0 m2) = 2.026 x 105 N
MANISH KUMAR
EXERCISE – 1 1.
2.
3.
4.
5. 6. 7.
8.
9.
10.
PHYSICS
GRAVIVATION
If a rock is brought from the surface of the moon. (a) its mass will change (b) Its weight will change but not mass (c) both mass and weight will change (d) its mass and weight both will remain same A body is weighed at the poles and than at the equator the weight. (a) at the equator will be greater then at the poles (b) at the pole will be greater than at the equator (c) at the poles will be equal to the weight at the equator (d) depends upon the object Consider a satellite going round the earth in a circular orbit. Which of the following statements is wrong ? (a) It is a freely falling body. (b) It is moving with constant speed. (c) It is acted upon by a force directed away from the centre of the earth which counterbalance the gravitational pull (d) Its angular momentum remains constant A missile is launched with a velocity less then the escape velocity. The sun of its kinl litic and potential energy is (a) Positive (b) negative (c) zero (d) may be positive or negative depending upon its initial velocity SI unit of g is (a) m2/s (b) s/m2 (c) m/s2 (d) m/s SI unit of G is (a) N2 – m2/kg (b) N - m2/kg (c) N – m/kg (d) N - m2/kg2 Choose the correct statement of the following : (a) All bodies repel each other in this universe (b) Our earth does not behave like a magnet. (c) Acceleration due to gravity is 8.9 m/s2 (d) All bodies fall at the same rate in vacuum. Maximum weight of the earth (a) at the centre of the earth (b) inside the earth (c) on the surface of the earth (d) above the surface of earth If the distance between two masses be doubled, then the force between them will become 1 1 (a) tomes (b) 4 time (c) times (d) 2 times 4 4 A body falls freely towards the earth with
MANISH KUMAR (a) uniform speed (d) uniform acceleration
PHYSICS (b) uniform velocity (d) none of these
MANISH KUMAR 11.
If the mass of a body is M on the surface of the earth, then its mass of surface of the moon will be M 6 Weight (a) is a vector quantity
(a)
12.
PHYSICS
(b) M
(c) M + 6
(d) zero
(b) of a body in interplanetary space is maximum (d) none of these
13.
(c) increases when the bodies go up The value of g near the earth’s surface is
(c) 9.8 m/s2
14.
(a) 8.9 m/s2 A geostationary satellite
(b) 8.9 m/s
(d) 9.8 m/s
(a) moves faster then the near earth satellite (b) has a time period less than of a near earth satellite (c) revolves about the polar axis (d) is stationary in space 15.
16.
17.
The force of gravitation between two bodies depend upon (a) their separation (b) gravitational constant (c) product of their masses (d) all of these When an object is thrown up, the force of gravity (a) acts in the direction of the motion (c) remains constant as the body moves up
(b) acts in the opposite direction of the motion (d) increases as the body moves up
The force of gravitational exists (a) everywhere in the universe
(b) at the surface of the earth only
18.
(c) inside the earth only 1 kg wt is equal to
(c) at the surface of the moon only (b) 9.80 dynes
(c) 98 dynes
(d) none of these
19.
(a) 9.8 N 1 kg wt is equal to
(c) 98 dynes
(d) none of these
20.
(a) 980 dynes (b) 9.80 dynes The value of G does not depend on (a) nature of the interacting bodies (b) size of the interacting bodies (c) mass of the interacting bodies (d) all of these
21.
The mass of the Jupiter is 1.9 x 1027 kg and that of sun is 1.99 x 1038 kg. The mean distance of the Jupiter from the sun is 7.8 x 1011 m. Speed of the Jupiter is (assuming that Jupiter moves in a circular orbit around the sun) (a) 1.304 x 104 m/sec
(b) 13.04 x 104 m/sec
MANISH KUMAR (c) 1.304 x 106 m/sec
PHYSICS (d) 1.304 x 102 m/sec
MANISH KUMAR 22.
PHYSICS
The acceleration due to gravity (a) has the same value everywhere in space (b) has the same value everywhere on the earth (c) varies with the latitude on the earth (d) is greater on the moon due to its smaller diameter
23.
SI unit of weight is (a) kg wt
24.
(c) g wt
(d) none of these
(c) 980 N
(d) none of these
Gravitational force which acts on 10 kg is (a) 9.8 N
25.
(b) N
(b)
1 N 9.8
Weight is (a) measured by a spring balance (b) measured by a balance (c) measured in kg (d) a scalar quantity
26.
A satellite which is geostationary in particular orbit is taken to another orbit. Its distance form the centre of earth in new obit is 2 times that of the earlier orbit. The time period in the second orbit is (a) 4.8 hours
27.
2 hours
(c) 24 hours
(d)24
2 hours
Two mater spheres of equal radius r are touching each other. The force of attraction F between them is (a) F r4
28.
(b) 48
(b) F r6
(c) F r2
(d) F
1 r2
A man weight 60 kg at earth’s surface. At whet height above the earth’s surface his weight becomes 30 kg
? (a) 1624 km 29.
(b) 2424 km
(c) 2624 km
(d) 2826 km
There are two bodies of masses 10 kg and 1000 kg separated by a distance 1 m. At what distance from the smaller body, intensity of gravitational field will be zero ? (a)
30.
16 m 24
1 m 10
(c)
1 m 11
(d)
10 m 11
At what height, is the value of g half that on the surface of earth ? (R = radius of the earth) (a) 0.414R
31.
(b)
(b) R
(c) 2R
(d) 3.5R
A planet of mass m moves around the sun of mass M in elliptical orbit. The maximum and minimum distance of the planet from the sun are r1 and r2 respectively. The time period after planet is proportional to
MANISH KUMAR
PHYSICS
(a) r13/2
32.
33.
(b) r23/2
(c)
(r1 r2 )3 / 2 2
(d)
(r1 r2 )3 / 2 2
On a planet (whose size the same and mass 4 time as that of the earth). The energy needed to lift a 2 kg. ma vertically upwards through 2 m distance on the planet is (g = 10 m/s2 on the surface of earth) (a) 16 Joules (b) 160 joules (c) 32 joules (d) 320 joules A satellite is revolving in a circular orbit a distance of 2620 km from the surface of the earth. The time period of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 x 1024 k.G= 6.67 x 10 N-m2 /kg2)
34.
(a) 2.35 hours (b) 23.5 hours (c) 3.25 hours (d) 32.5 hours On the surface of the earth, force of gravitational attraction between two masses kept at distance d apart it 6 Newton’s. If these two masses are taken to the surface of the moon kept at the same distance d, the force between them will be (a) 1 N
(b) 36 N
(c)
1 N 6
(d) 6 N
35.
Time period of simple pendulum in a satellite is
36.
(a) infinite (b) zero (c) 2 sec (d) cannot be calculated A ball is dropped from a spacecraft revolving around the earth at height of 120 km. What will happed to the ball ? (a) It will continue to move with the same speed along the original orbit of spacecraft (b) It will move with the same speed, tangentially to the spacecraft (c) It will fall down to the earth gradually (d) It will go very far in space
ANSWER KEY Que. 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
B
B
C
B
B
D
D
C
A
C
B
A
C
C
D
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
Que. 16
MANISH KUMAR Ans.
B
Que. 31 Ans.
C
PHYSICS
A
A
A
D
A
32
33
34
35
36
B
A
D
A
A
C
B
A
A
D
A
C
C
A
MANISH KUMAR
PHYSICS
EXERICSE -2
GRAVITATION
1. 2. 3. 4. 5.
6. 7. 8. 9. 10. 11.
12. 13. 14. 15. 16. 17. 18.
19. 20.
What is the gravitational of a spaceship at a distance equal to two Earth’s radius from the centre of the Earth ? A body on diff 49m high drops a stone. One second later, he throw a second stone after the ftrst. They both hit the ground at the same time. With what speed did he the second stone ? A stone drops from the edge of the root. It passes a window 2 m high in 0.1s How far is the roof above the top of the window ? A particle is dropped from a town 180 m high. How long does it take reach the ground ? what is the velocity when it tourhes the ground ? Take g = 10 m/s2 To estimate the height of a bridge over a river, a stone is dropped freely on the river from the bridge . The stone takes 2 s to touch the water surface in the river. Calculate the height to the bridges from the water level. Take g 9.8m/s2 How much would a 70 kg man weigh on moon ? What will be his mass on Earth and Moon? Given g on Moon = 1.7 m/s2 A body has a weight of 10 kg on the surface of Earth. What will be its mass and weight when taken to the centre of Earth ? A force of 2 kg wt acts on the surface A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration ? A man weight 600 N on the earth. What is its mass ? Take g = 10 m/s2. If he were take on moon, his weight would be 100 N. What is his mass on moon ? What is acceleration due to gravity on Moon ? A car falls off a ledge and drops to the ground on 0.5 s Let = 10 m. (for simplifying the calculations) (i) When is its speed on striking the ground ? (ii) What is its average speed during 0.5s ? (iii) How high is the ledge from the ground ? An object is thrown vertically upwards and rises to a height of 10 m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken object to reach the highest point. Mass of and object is 10 kg. What is its weight on Earth ? An object weight 10 N when measured on the surface of the Earth. What would be its weight when measure on the surface of moon ? Calculate the value of acceleration due to gravity on Moon. Given mass of Moon = 7.4 x 1022 Kg. radius of Moon =1740 km. Suppose a planet exists whose mass and radius both are half those of Earth. Calculate the acceleration due gravity it the surface of this planet. A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall ? Take g = 9.8 m/s2. A block of wood it kept on a tabletop. The mass of wooden block is 5 kg and its and its dimensions are 40 cm x 20 cm x 10 cm. Find the pressure exerted by the wooden block on the table to if its is made to lie on the table to with its sides of dimensions (a) 20 cm x 10 cm and (b) 40 cm x 20 cm. Relative density of silver is 10.8 The density of water is 103 m-3 What is density of silver in SI unit ? How does the force of gravitation between two objects change when the distance between then is reduced half ?
MANISH KUMAR 21. 22. 23. 24.
25. 26. 27
PHYSICS
Gravitational force acts on all object in proportion to their masses. When then. A heavy object does not fall faster then a light object What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface ? (Mass of the earth is 6 x 1024 kg ad radius of the earth is 6.4 x 106 m.) 1 Gravitational force on the surface of the moon is only as strong gravitational force on the earth. What the 6 weight in newtons of a 10kg object on the moon and on the earth ? A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth. A stone is released from the top of a tower or height 19.6m Calculate it s final velocity just before touching the ground. A stone is thrown upward with an initial velocity of 40 m/s Taking g = 10 m/s2, find the maximum height reached by the stone. What is tenet displacement and the total distance covered by the stone ? Calculate force of gravitational between the earth and the sun. given that the mass of the mass of the earth = 6 x 1024 kg and of the sun = 2 x 1030 kg . The average distance between the two is 1.5 x 1011 m.
ANSWER KEY 1. 4. 7. 10. 12. 15. 18. 19. 22. 23. 24. 25. 26. 27.
2-45 m/s2 2. 12.1 M/S 3. 19.4 m 6 S, 60 m/s 5. 19.6 m 6. 119 N, 70 kg, 70 kg 10 kg, Zero 8. 4 m/s2 9. 1 kg, 20 m/s2 2 60 kg, 60 kg, 1.67 m/s 11. (i) 5 m/s (ii) 2.5 m/s (iii) 1.25 m) 14 m/s, 1.43 s 13. 98 N 14. 1.67 N 1.63 m/s2 16. 19.6 m/s2 17. 11.48 m (i) 2459 Nm2 (ii) 6125 Nm-2 10.8 x 103 kg m-3 20. 4F 21. anet = g 9.8 N Weight on earth is 98 N and on moon is 16.3 N Maximum height is 122.5 m and total time is 5s + 5 = 10s 19.6 m/s Maximum height = 80m, Net displacement = 0, Total distance covered = 160 m Gravitational force = 53.36 x 1032 N
MANISH KUMAR
PHYSICS
MANISH KUMAR
PHYSICS
Important Notes
