patel (rpp467) – Quest #8 Gravity (part 2/3) – ha – (51122013) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two spheres have equal densities and are subject only to their mutual gravitational attraction.
Which quantity must have the same magnitude for both spheres? 1. velocity 2. acceleration 3. kinetic energy
003 (part 1 of 2) 10.0 points A coordinate system (in meters) is constructed on the surface of a pool table, and three objects are placed on the coordinate system as follows: a 0.8 kg object at the origin, a 2.9 kg object at (0 m,1.7 m), and a 4.3 kg object at (4.4 m,0 m). Find the resultant gravitational force exerted on the object at the origin by the other two objects. The universal gravitational constant is 6.672 × 10−11 N · m2 /kg2 . Correct answer: 5.4857 × 10−11 N. Explanation: Let : m1 = 0.8 kg , m2 = 2.9 kg , r = 1.7 m , and G = 6.672 × 10−11 N · m2 /kg2 .
4. displacement from the center of mass 5. gravitational force correct Explanation: Two spheres with the same density have different masses due to their relative sizes. Using Newton’s third law, F~1 = −F~2 . All of the other quantities (acceleration, velocity, kinetic energy, and displacement from the center of mass) have different magnitudes because the two spheres have different masses. 002 10.0 points The strength of the gravitational force between objects depends on 1. the mass of each object. 2. the distance between the objects. 3. neither the masses nor the distance between them. 4. the mass of each object and the distance between them. correct Explanation:
1
2.9 kg 1.7 m
Fy
θ
.8 kg Fx
4.3 kg 4.4 m
The force exerted on the 0.8 kg mass by the 2.9 kg mass is in the positive y direction and given by G m1 m2 Fy = r2 = 6.672 × 10−11 N · m2 /kg2 (0.8 kg) (2.9 kg) × (1.7 m)2 = 5.35607 × 10−11 N . The force exerted on the 0.8 kg object by the 4.3 kg mass is in the positive x direction and is G m1 m2 Fx = r2 = 6.672 × 10−11 N · m2 /kg2 (0.8 kg) (4.3 kg) × (4.4 m)2 = 1.18552 × 10−11 N .
patel (rpp467) – Quest #8 Gravity (part 2/3) – ha – (51122013) 2 s The resultant force is 4 π 2 r3 T = 2 2 2 FR = Fy + Fx G Ms s −11 2 = (5.35607 × 10 N) 4 π 2 (8.25 × 1011 m)3 −11 T = + (1.18552 × 10 N)2 6.6726 × 10−11 N · m2 /kg2 −21 2 r = 3.00929 × 10 N , so p 1 × FR = 3.00929 × 10−21 N2 1.98 × 1030 kg −11 1d 1y 1h = 5.4857 × 10 N . × × × 3600 s 24 h 365 d = 12.9889 y . 004 (part 2 of 2) 10.0 points At what angle does it act with respect to the positive x axis? Let counterclockwise be positive, within the limits of −180◦ to 180◦ . 006 10.0 points Given: G = 6.67259 × 10−11 N m2 /kg2 Correct answer: 77.5193◦. A planet is orbiting a star. Calculate the mass of the star using the fact Explanation: that the period of the planet is 3.16 × 107 s and its distance from the star is 2.31 × 1011 m. Fy tan θ = Fx Correct answer: 7.30343 × 1030 kg. F y θ = tan−1 Explanation: Fx −11 From Kepler’s third law, we have 5.35607 × 10 N = tan−1 1.18552 × 10−11 N 4 π 2 r3 ◦ M= = 77.5193 G T2 4 π2 above the +x-axis. = (6.67259 × 10−11 N m2 /kg2 ) 005 10.0 points (2.31 × 1011 m)3 × The asteroid Hektor, discovered in 1907, is in (3.16 × 107 s)2 a nearly circular orbit of radius 5.5 AU about = 7.30343 × 1030 kg . the sun, which has a mass of 1.98 × 1030 kg. Determine the period of this asteroid. Correct answer: 12.9889 y.
007 10.0 points A 8.69479 kg mass experiences a gravitational force of (144.776 N)ˆı at some point P. What is the gravitational field at that point?
Explanation: 1.5 × 1011 m , AU = 8.25 × 1011 m , G = 6.6726 × 10−11 N · m2 /kg2 , Ms = 1.98 × 1030 kg .
Let : r = 5.5 AU ×
Correct answer: 16.6509 N/kg. and
Explanation:
Applying Kepler’s third law, T 2 = C r3 =
4 π2 3 r G Ms
Let : m = 8.69479 kg and fˆ = (144.776 N)ˆı .
patel (rpp467) – Quest #8 Gravity (part 2/3) – ha – (51122013) The gravitational field at a point in space is ~g =
~f (144.776 N)ˆı = = (16.6509 N/kg)ˆı . m 8.69479 kg
008 10.0 points A thin spherical shell has a radius of 5 m and a mass of 353 kg, and its center is located at the origin of a coordinate system. Another thin spherical shell with a radius of 3 m and mass 159 kg is inside the larger shell with its center at 0.31 m on the x axis. What is the gravitational force of attraction between the two shells? Correct answer: 0 N.
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The mass of a differential element of the sphere is C dm = ρ dV = (4 π r 2 dr) . r Z Z r M = dm = 4 π C R dR , 0 r 1 2 = 4π C R 2 0 = 2 π C r2 = (23.3928 m2 ) π C M C= (23.3928 m2 ) π 1011 kg = 13.7569 kg/m2 . = (23.3928 m2 ) π
010 (part 2 of 3) 10.0 points Find the gravitational field 5.03 m from the center of the sphere.
Explanation: Let : r1 = 5 m , m1 = 353 kg , r2 = 3 m , m2 = 159 kg , d = 0.31 m .
Correct answer: 2.66631 × 10−9 N/kg. Explanation: and
Let : r = 5.03 m . g=
The gravitational field inside the 5 m shell due to that shell is zero, so it exerts no force on the 3 m shell, and by Newton’s third law, that shell exerts no force on the large shell. Thus the gravitational attraction is F =0N .
(6.6726 × 10−11 N · m2 /kg2 ) (1011 kg) (5.03 m)2
= 2.66631 × 10−9 N/kg . 011 (part 3 of 3) 10.0 points Find the gravitational field 1.13 m from the centre of the sphere. Correct answer: 5.76759 × 10−9 N/kg.
009 (part 1 of 3) 10.0 points C The density of a sphere is given by ρ(r) = . r The sphere has a radius of 3.42 m and a mass of 1011 kg. Find the constant C. Correct answer: 13.7569 kg/m2 . Explanation: r = 3.42 m and M = 1011 kg .
Explanation: Let : R = 1.13 m . The gravitational field inside the sphere is GM g= R2 2 π G C R2 = R2 = 2πC G = 2π(13.7569 kg/m2 ) × (6.6726 × 10−11 N · m2 /kg2 ) = 5.76759 × 10−9 N/kg .
patel (rpp467) – Quest #8 Gravity (part 2/3) – ha – (51122013) 012 10.0 points An object is dropped from rest from a height 3.4 × 106 m above the surface of the earth. If there is no air resistance, what is its speed when it strikes the earth? The acceleration of gravity is 9.81 m/s2 and the radius of the Earth is 6.37 × 106 m.
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An object is projected upward from the surface of the earth with an initial speed of 2 km/s. The acceleration of gravity is 9.81 m/s2 . Find the maximum height it reaches. Correct answer: 210.614 km. Explanation:
Correct answer: 6.59495 km/s. Explanation: Let :
h = 3.4 × 106 m , g = 9.81 m/s2 , and r = 6.37 × 106 m .
G ME and the potential energy at a 2 RE distance r from the surface of the earth is
Let : RE = 6.37 × 106 m , vi = 2 km/s = 2000 m/s , g = 9.81 m/s2 .
G ME and the potential energy at a 2 RE distance r from the surface of the earth is
g =
g =
U (r) = −
G ME m . r
Using conservation of energy to relate the initial potential energy of the system to its energy as the object is about to strike the earth (Ki = 0),
U (r) = −
G ME m r
Using conservation of energy to relate the initial potential energy of the system to its energy at its maximum height (Kf = 0), K f − K i + Uf − Ui = 0 −K(RE ) + U (RE + h) − U (RE ) = 0 GME m GME m 1 − =0 − mvi2 + 2 RE + h RE
K f + Uf − Ui = 0 (RE ) + U (RE ) − U (RE + h) = 0 1 G ME m G ME m m v2 − + =0 2 RE RE + h s G ME G ME v= 2 − RE RE + h s h = 2 g RE RE + h s 2 (9.81 m/s2 ) (6.37 × 106 m) = 3.4 × 106 m + 6.37 × 106 m p 1 km × 3.4 × 106 m · 1000 m = 6.59495 km/s .
and
G ME h RE 1 2 vi = · 2 RE (RE + h) RE 2 G ME h RE vi2 = · 2 RE + h RE 2 g h RE vi2 = RE + h 2 (RE + h) vi = 2 g h RE −v 2 RE h= 2 i vi − 2 g RE Since vi2 − 2 g RE = (2000 m/s)2 h − 2 (9.81 m/s2 )
×(6.37 × 106 m)
013
10.0 points
= −1.20979 × 108 ,
i
patel (rpp467) – Quest #8 Gravity (part 2/3) – ha – (51122013)
(2000 m/s)2 (6.37 × 106 m) 1 km h=− · −1.20979 × 108 1000 m = 210.614 km .
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is 6.67259 × 10−11 N · m2 /kg2 , the mass of the earth is 5.98 × 1024 kg and its radius is 6.37 × 106 m . Correct answer: 7792 m/s. Explanation:
014 10.0 points The planet Saturn has a mass 95.2 times that of the earth and a radius 9.47 times that of the earth. Find the escape speed for objects on the surface of Saturn.
G = 6.67259 × 10−11 N · m2 /kg2 , M = 5.98 × 1024 kg , m = 149 kg , R = 6.37 × 106 m , and h = 202 km = 2.02 × 105 m .
Let :
Correct answer: 35.5109 km/s. Explanation:
The radius of the satellite’s orbit is r = R + h = 6.37 × 106 m + 2.02 × 105 m = 6.572 × 106 m .
Let : MS = 95.2 ME and RS = 9.47 RE . The escape speed from Saturn is s 2 G MS ve.S = RS and the escape speed from Earth is s 2 G ME ve.E = . RE
The gravitational force provides the centripetal acceleration, so (1)
(2)
Dividing, ve.S = ve.E
s
RE MS · RS ME
ve.S =
s
RE MS · ve.E RS ME
=
r
95.2 1 · (11.2 km/s) 9.47 1
= 35.5109 km/s . 015 (part 1 of 3) 10.0 points A satellite of mass 149 kg is launched from a site on the Equator into an orbit at 202 km above Earth’s surface. If the orbit is circular, what is the satellite’s speed in orbit? The gravitational constant
m
v2 r
GM m r2 r GM v= q r =
=
6.67259 × 10−11 N · m2 /kg2 s 5.98 × 1024 kg × 6.572 × 106 m
= 7792 m/s .
016 (part 2 of 3) 10.0 points What is the orbital period of this satellite? Correct answer: 1.47206 h. Explanation: The period of the orbital motion is 2πr v 2 π (6.572 × 106 m) 1 h = 7792 m/s 3600 s
T =
= 1.47206 h .
patel (rpp467) – Quest #8 Gravity (part 2/3) – ha – (51122013) 017 (part 3 of 3) 10.0 points What is the minimum energy necessary to place this satellite in orbit, assuming no air friction? Correct answer: 4.79418 × 109 J. Explanation: Assuming the satellite is launched from a point on the equator of the Earth, its initial speed is the rotational speed of the launch point, or 2πR t 2 π 6.572 × 106 m 1 d h = 1d 24 h 3600 s = 463.239 m/s .
vi =
The work-kinetic energy theorem gives the energy input required to place the satellite in orbit as W = (K + P )f − (K + P )i 1 GM m 2 = m vt − 2 r 1 GM m 2 − m vi − 2 R 2 2 vt − vi 1 1 =m + GM m − 2 R r = (149 kg) (7792 m/s)2 − (463.239 m/s)2 × 2 −11 + 6.67259 × 10 N · m2 /kg2 × 5.98 × 1024 kg (149 kg) 1 1 × − 6.37 × 106 m 6.572 × 106 m
= 4.79418 × 109 J .
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