• Study Resource
• Explore

Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Gravity wikipedia, lookup

Inertia wikipedia, lookup

Relativistic mechanics wikipedia, lookup

Seismometer wikipedia, lookup

Center of mass wikipedia, lookup

Electromagnetic mass wikipedia, lookup

Modified Newtonian dynamics wikipedia, lookup

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
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
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 
```
Related documents