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What multiple of the energy needed to escape from Earth givesthe energy needed to escape from (a) the Moon and (b) Jupiter?

Short Answer

Expert verified
  1. The ratio of energy needed to escape from the surface of the moon to the energy needed to escape from the earth is0.045 .
  2. The ratio of energy needed to escape from the surface of Jupiter to energy needed to escape from the earth is 28.5.

Step by step solution

01

Step 1: Given

Mass of earth,5.95×1024kg

Mass of moon,7.36×1022kg

Mass of Jupiter,1.89×1027kg

The radius of the earth,6.37×106m

The radius of the moon,1.74×106m

The radius of Jupiter, 6.99×107m

02

Determining the concept

The amount of energy needed to escape from the surface is the same as its gravitational potential energy at the original position.

The formula is as follows:

K=GMmR

where, M, and m are masses, R is the radius, G is the gravitational constant and K is energy.

03

(a) Determining the ratio of energy needed to escape from the surface of the moon to energy needed to escape from the earth

The gravitational potential energy is written as,

ΔU=-GMmR

The amount of energy required to escape from the surface would be the absolute value of gravitational potential energy, so,

K=GMmR

The ratio of energy needed to escape from the surface of the earth to energy needed to escape from the moon,

KMKE=MMREMERM=7.36×1022kg×6.37×106m5.95×1024kg×1.74×106m=0.045

Using the values from Appendix C,

KMKE=0.045

Hence, the ratio of energy needed to escape from the surface of the moon to the energy needed to escape from the earth is 0.045.

04

(b) Determining the ratio of energy needed to escape from the surface of Jupiter to energy needed to escape from earth

Using the values from Appendix C,

KJKE=MJREMERJ=1.89×1027kg×6.37×106m5.95×1024kg×6.99×107m=28.5

Hence, the ratio of energy needed to escape from the surface of Jupiter to energy needed to escape from the earth is 28.5.

Therefore, the amount of energy needed to escape from the surface is the same as its gravitational potential energy at the original position. Using this concept, the given ratios can be found.

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