QUESTION 1
Part 1
Engineers wish to launch a satellite from the surface of the Moon. What is the minimum speed the satellite must have to escape the Moon’s gravity – that is, what is the escape velocity at the surface of the Moon? The Moon has a mass of 7.3x10^22 kg and a radius of 1.7x10^6 m.
a. 
1700 m/s 

b. 
5.7x10^6 m/s 

c. 
It depends on the mass of the satellite. 

d. 
2400 m/s 
Part 2
The satellite is guided to a circular orbit about the Sun at the same distance from the Sun as the Earth (1.5x10^11 m), but far away from the Earth (so only the Sun’s gravity is important). What is the satellite’s gravitational acceleration due to the Sun, which has a mass of 2.0x10^30 kg?
a. 
5.9 mm/s^2 

b. 
9.8 m/s^2 

c. 
8.9x10^8 m/s^2 

d. 
It depends on the mass of the satellite. 
Part 3
The satellite is guided to a circular orbit about the Sun at the same distance from the Sun as the Earth (1.5x10^11 m), but far away from the Earth (so only the Sun’s gravity is important). What is the satellite’s speed?
a. 
3.0x10^4 m/s 

b. 
8.9x10^8 m/s 

c. 
5.9 mm/s 

d. 
It depends on the mass of the satellite. 
Part 4
The satellite is guided to a circular orbit about the Sun at the same distance from the Sun as the Earth (1.5x10^11 m), but far away from the Earth (so only the Sun’s gravity is important). What is the satellite’s gravitational potential energy if it has a mass of 200 kg? The Sun has a mass of 2.0x10^30 kg.
a. 
1.8x10^11 J 

b. 
8.9x10^8 J 

c. 
2.9x10^14 J 

d. 
1.2 J 
Part 5
The satellite is guided to a circular orbit about the Sun at the same distance from the Sun as the Earth (1.5x10^11 m), but far away from the Earth (so only the Sun’s gravity is important). The satellite is the stopped using orbital thrusters and allowed to fall to the Sun. What is the satellite’s speed when it reaches the surface of the Sun? The Sun has a mass of 2.0x10^30 kg and a radius of 7.0x10^8 m.
Hint: Use energy conservation. First consider the potential energy when it is at the Earth’s orbit, then equate this to the sum of the potential energy at the surface of the sun and the kinetic energy at the surface of the sun.
a. 
4.4x10^5 m/s 

b. 
6.2x10^5 m/s 

c. 
It will be going infinitely fast. 

d. 
3.8x10^11 m/s 
part 1)
We know that escape velocity
here G=6.67*10^{11} , M=7.3*10^{22} kg , 1.7*10^{6} m
V=23.93*10^{2} m/s
V=2400 m/s Answer
part 2) As you know gravitational force is equal to
F=mg=GMm/R^{2}
g=GM/R^{2}
=6.67*10^{11}*2*10^{30}/(1.5*10^{11})^{2}
g=5.93*10^{3} m/s^{2}
g=5.9 mm/s^{2} Answer
part 3) As you know gravitational force = centripetal force
GMm/r^{2} =mv^{2}/r So
v^{2} =GM/r
here M=2*10^{3}^{0} kg , r=1.5*10^{11} m
v^{2} =6.67*10^{11}*2*10^{30}/1.5*10^{11}
v^{2} =8.893*10^{8}
v=2.98*10^{4} m/s
v=3.0*10^{4} m/s Answer
part 4) We know that potential energy of the satellite
U=GMm/r
U=6.67*10^{11}*2*10^{30}*200/1.5*10^{11}
U=1778.67*10^{8} J
U=1.8*10^{11} J Answer
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