Download Assignment 8

INNOVA JUNIOR COLLEGE
JC1 Physics
Tutorial 8: Gravitational Field
25
1. The acceleration of free fall on the surface of the Earth is 6 times its value on the surface of
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the Moon. The mean density of the Earth is
times the mean density of the Moon. Given
3
r
that rE is the radius of the Earth and rM is the radius of the Moon, show that the value of E
rM
is 3.6.
[2]
aE = 6aM
gE = 6gM
Since g 
GM
r2
4 3
r
3

r2
4
 G  r
3
4
4
GE  rE  6GM  rM
3
3
rE
M
6
rM
E
3
 6
5
= 3.6
G
[1]
[1]
2. (a) The Earth may be assumed to be an isolated sphere of radius 6.4 × 103 km with its mass
of 6.0 × 1024 kg concentrated at its centre. An object is projected vertically from the
surface of the Earth so that it reaches an altitude of 1.3 × 104 km.
Calculate, for this object,
(i) the change in gravitational potential,
[2]
(ii) the speed of projection from the Earth’s surface, assuming air resistance is negligible.
[3]
(b) Suggest why the equation
v2 = u2 + 2as
is not appropriate for the calculation in (a) (ii)
[1]
(a) (i)   f  i
 GM   GM 
   
  

 rf   ri 

1
1
 

  6.67  10 11  6.4  106  

7 
6 
 1.94  10   6.4  10  
7
-1
= 4.19 × 10 J kg
-1-
[1]
[1]
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(ii) Assuming air resistance is negligible,
GPEi + KEi = GPEf + KEf
mi + ½ mvi2 = mf + 0
½ mv2 = mf - mI
v  2
[1]
[1]
 2  4.19  107
= 9.15 × 103 m s-1
[1]
(b) The resultant force, comprising of gravitation force is not constant from the surface of the
Earth till it reaches the altitude of 1.3 × 104 km since it follows the inverse square law.
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(a) The Moon may be considered to travel about the Earth in a circular orbit of radius
3.82 × 108 m and period 2.36 × 106 s. Determine the acceleration of the Moon.
[2]
(b) Why does the Moon not fall and hit the Earth?
[2]
(c) By considering the acceleration of free fall at the Earth’s surface, show that the
magnitude of the Moon’s acceleration is consistent with Newton’s inverse square law of
gravitation.
[3]
[Radius of the Earth = 6.36 × 106 m]
 2 
(a) ac  r  2  r 

T 
2
2


ac  3.82  10  
6 
 2.36  10 
-3
ac = 2.71 × 10 m s-2
2
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[1]
[1]
(b) The resultant force (i.e. gravitational force) acting on Moon is acting perpendicular to its
velocity [1]. Thus this gravitational force is responsible for the uniform circular motion, i.e.
changing the direction of the velocity [1].
(c) If
1
r2
k
a 2
r
2
ar  k
a

For surface of the Earth, ks  9.81 6.36  106
For Moon’s orbit,


2
 3.97  1014
[1]

[1]
km  2.708  103  3.82  108
2
 3.95  1014
Since the constants k are similar for the surface of the Earth and Moon’s orbit [1], the
acceleration is consistent with Newton’s inverse square law of gravitation.
4. A satellite of mass m orbits a planet of mass M and radius Rp. The radius of the orbit is R.
The satellite and the planet may be considered to be point masses with their masses
concentrated at their centres. They may be assumed to be isolated in space.
(a) (i) Derive an expression, in terms of M, m and R, for the kinetic energy of the satellite.
Explain your working.
[2]
(ii) Show that, for the satellite in orbit, the ratio
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gravitational potential energy of satellite
is equal to -2.
kinetic energy of satellite
(a) (i) Fc = mac
GMm mv 2
[1]

r
r2
GMm
mv 2 
r
GMm
[1]
KE  ½mv 2 
2r
(ii) GPE  
GMm
r
[1]
[1]
GPE
KE
GMm

r

GMm
2r
 2
[0]
(b) The variation with orbital radius R of the gravitational potential energy of the satellite is
shown in Fig 1.1
Shape of KE of satellite
1. Mirror image about the
x-axis [1]
2. × ½ the value of GPE [1]
Fig 1.1
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(i) On Fig 1.1, draw the variation with orbital radius of the kinetic energy of the satellite.
Your line should extend from R = 1.5 Rp to R = 4 Rp
[2]
(ii) The mass m of the satellite is 1600 kg.
The radius of the orbit of the satellite is changed from R = 4 Rp to R = 2 Rp.
Use Fig 1.1 to determine the change in orbital speed of the satellite.
[5]
(ii) When R = 4Rp, KE = 1.25 × 109 J
½ (1600) v2 = 1.25 × 109
v = 1.250 × 103 m s-1
[1]
When R = 2Rp, KE = 2.50 × 109 J
½ (1600) v’2 = 2.50 × 109
v' = 1.717 × 103 m s-1
[1]
Change in orbital speed, Δv = v’ – v = 517 m s-1
[1]
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[1]
[1]
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