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Transcript
Q08. Gravity
1. Suppose you have a pendulum clock which keeps correct
time on Earth (acceleration due to gravity = 9.8 m/s2).
Without changing the clock, you take it to the Moon
(acceleration due to gravity = 1.6 m/s2).
For every hour
interval (on Earth) the Moon clock will record:
1.
(9.8/1.6) h
2.
1h
3.
9.8 / 1.6 h
4.
(1.6/9.8) h
5.
1.6 / 9.8 h
2. The mass density of a certain planet has spherical symmetry
but varies in such a way that the mass inside every spherical
surface with center at the center of the planet is proportional to
the radius of the surface.
If r is the distance from the center
of the planet to a point mass inside the planet, the gravitational
force on the mass is :
1.
not dependent on r
2.
proportional to r2
3.
proportional to r
4.
proportional to 1/ r
5.
proportional to 1/ r2
Mass inside radius R :
M  R      r  4 r d r  R
R
2

0

1
r2
Force due to mass shell beyond m vanishes.
Force due to mass sphere beneath m is the same as if it’s a
point mass M(r) placed at the center.
m M r
f G
r2

r
1

r2
r
3. Each of the four corners of a square with edge a is occupied by a
point mass m. There is a fifth mass, also m, at the center of the
square. To remove the mass from the center to a point far away
the work that must be done by an external agent is given by :
1.
4Gm2/a
2.
–4Gm2/a
3.
(42) Gm2/a
4.
–(42) Gm2/a
5.
4Gm2/a2
Potential energy of m at center of square is :



m2 
U  0   4   G

2 

a

2 

U     0 requires work
To bring it to infinity, where
W0ext
  W0 
8 Gm 2
Gm 2

 4 2
a
2 a
 U0
 U     U  0
Gm 2
4 2
a
4. A projectile is fired straight upward from Earth's surface with a
speed that is half the escape speed.
If R is the radius of Earth,
the highest altitude reached, measured from the surface, is
1.
R/4
2.
R/3
3.
R/2
4.
R
5.
2R
1 v 
K  m  esc 
2  2 
By definition:
2
1 2
M
vesc  G E
2
R
Maximum height H is given by:
E  K  U  R  U  R  H 
1 v 
K  m  esc 
2  2 
2
1
1

 GmM E  
 
 RH R
1
1
1
 
4R R R  H
H

R
3
1
3

R  H 4R
( Energy conservation )
5. An object is dropped from an altitude of one Earth radius
above Earth's surface. If M is the mass of Earth and R is its
radius the speed of the object just before it hits Earth is given
by :
1.
GM / R
2.
GM / 2R
3.
2GM / R
4.
GM / R 2
5.
GM / 2 R 2
Energy conservation :
E  U  2R   K  U  R 
1 2
1
 1
mv  GmM  
 
2
 2R R 
v
GM
R
1 GmM

2 R
6. A planet in another solar system orbits a star with a mass of 4.0 
1030 kg.
At one point in its orbit it is 250  106 km from the
star and is moving at 35 km/s.
Take the universal gravitational
constant to be 6.67  10–11 m2/s2  kg and calculate the semimajor
axis of the planet's orbit.
The result is :
1.
79  106 km
2.
160  106 km
3.
290  106 km
4.
320  106 km
5.
590  106 km
Energy conservation :
30
1

4.0

10
2


3
11
E  K  U  M   35  10    6.67  10 
9 
2
250

10


 4.547  108 M
Using
E  G
M MS
2a
( M = mass of planet )
( MS = mass of star, a = semimajor axis ) , we have
6.67  10  4.0  10 

a
2   4.547  10 
11
30
8
 2.93  1011 m  290  106 km
7. A spaceship is returning to Earth with its engine turned off.
Consider only the gravitational field of Earth and let M be the
mass of Earth, m be the mass of the spaceship, and R be the
distance from the center of Earth.
In moving from position 1
to position 2 the kinetic energy of the spaceship increases by :
1.
2.
3.
4.
5.
 1
1 
GMm  2  2 
 R 2 R1 


 1
1 
GMm  2  2 
 R1 R 2 


R R
GMm 1 2 22
R1 R 2
R1  R2
GMm
R1 R2
GMm
R 2  R1
R1 R2
Energy conservation :
E  K1  U1  K2  U 2
K2  K1  U1  U 2
 1
1 
 GMm    
 R1 R2 
 R1  R2 
 GMm 

R
R
 1 2 