Download Gravity - Pulling it all Together

Radius of Earth = 6.4 E 6m
Mass of Moon = 7.4 E 22 kg
Distance from Earth to Moon = 3.8 E8 m
Mass of Sun = 2.0 E 30 kg
Distance from Earth to Sun = 1.5 E 11m
1. If the mass of Earth were to be doubled at the same time
the radius was doubled, the free-fall acceleration would
(a) increase, (b) decrease, (c) stay the same?
Fg 
Gm(2)m E
(2) 2 R 2
Why?
(B) decrease, because the radius is squared
2. Why does an astronaut in a spacecraft orbiting Earth


Mass of Earth = 6.0 E 24 kg
experience a feeling of weightlessness?
He is in constant free fall around the Earth
3. Calculate the effective value of g on the top of Mount
Everest, 8850 m above seal level. (9.74 m/s2)
Gmm E
R2
Gmm E
mg 
R2
Fg 

(6.67 1011 )(6.0 10 24 )
g
 9.74m / s 2
6
2
(6.4 10  8850)
4. A 50.0kg person and a 75.0 kg person are sitting on a
bench 50 cm apart. Calculate the magnitude of the
gravitational force each exerts on the other. (1.0x10-6 N)
Gm1m 2 (6.67 1011 )(50kg)(75kg)
6
Fg 


1.0
10
N
R2
0.50m 2



5. Most of the stars in the galaxy of which the sun is a
member (the “Milky Way”) are concentrated in an assembly
about 100,000 light-years across whose shape is roughly
that of a fried egg. The sun is about 30,000 light-years
from the center of the galaxy, and revolves around it with
a period of about 2x108 years. A reasonable estimate for
the mass of the galaxy may be obtained by considering this
mass to be concentrated at the galactic center with the sun
revolving around it like a planet around the sun.
a. On this basis, calculate the mass of the galaxy.
(1
7
15
year = 3.15x10 sec., 1 light year = 9.46x10 m)
(3.4x1041 kg)
9.46 1015 m
r  30,000lys 
 2.84 10 20 m
1lyr
3.15 10 7 sec
8
T  2.0 10 yrs 
 6.311015 sec
1yr
Fc  Fg
m s 4 2 r Gm sm g

T2
R2
4 2 r 3
4 2 (2.84 10 20 ) 3
mg 

 3.4 10 41 kg
2
11
15 2
GT
(6.67 10 )(6.32 10 )
b. How many stars having the mass of the sun is this
equivalent to? (1.7x1011 suns)
mg 3.4 10 41
# suns 

1.7 1011 suns
30
ms 2.0 10
6. During a solar eclipse the moon comes between the Earth and
the Sun.
a. Find the net force on the Moon due to the gravitational
attraction of both the Earth and the Sun when they are
aligned Earth, Moon, Sun. (2.4x1020 N towards Sun)

Gm E m M (6.67 1011 )(6.0 10 24 )(7.4 10 22 )
20
FEM 


2.05
10
N
R2
(3.8 10 8 ) 2
Gm S m M (6.67 1011 )(2.0 10 30 )(7.4 10 22 )
FSM 

 4.4 10 20 N 
2
11 2
R
(1.496210 )
Fnet  2.05 10 20  4.4 10 20  2.4 10 20 towardsSun

b. One fine day, the Star Ship Enterprise decides to tow
the Earth out of its orbit and into deep space. Will
our Moon come with us? Explain.



No, the attraction between the moon and the sun is twice as
strong as the attraction between the Earth and the moon.
7. Some communication and weather satellites are launched into
circular orbits above the Earth’s equator so they are
synchronous with the Earth’s orbit.
That is they remain
“fixed” or “hover” over one point on the equator.
At what
altitude are these geosynchronous satellites? (3.6 E4 km)
T
24hrs 3600sec

 86400sec
1hr
Fc  Fg
m s 4 2 r Gm sm E

2
T
R2
Gm ET 2 (6.67 1011 )(6.0 10 24 )(86400) 2
3
R 

4 2GT 2
4 2
R  3 7.56710 22  42297524m
42297524m  RE  y  6.4 10 6  y
y  3.6 10 7 m  3.6 10 4 km