Question
Determine the escape velocity of a rocket on the far side of a moon of a planet. The radius of the moon is 2.64×106m and its mass is 1.495×1023. The mass of the planet is 1.9×1027kg, and the distance between planet and the moon is 1.071×109m. Include the gravitational effect of planet and neglect the motion of the planet and the moon as they rotate about their CM.
Answer: Option D
:
D
Total potential energy of the rocket is
U=−G[Mpm(d+Rm)+MmmRm]
If ve is the escape velocity,we can write
12mv2e=−U
v2e=2G(Mp(d+Rm)+MmRm)
= 2×6.67×10−11(1.90×10271.071×109+2.64×106+1.495×10232.64×106)
= 2.436×108
ve=1.560×104ms−1
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:
D
Total potential energy of the rocket is
U=−G[Mpm(d+Rm)+MmmRm]
If ve is the escape velocity,we can write
12mv2e=−U
v2e=2G(Mp(d+Rm)+MmRm)
= 2×6.67×10−11(1.90×10271.071×109+2.64×106+1.495×10232.64×106)
= 2.436×108
ve=1.560×104ms−1
Was this answer helpful ?
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