Q08._Gravity-Ans
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Transcript Q08._Gravity-Ans
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.
(42) Gm2/a
4.
–(42) 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
Gm2
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
RH 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