Question

A man weighs 750 N on the surface of the earth. What would be his weight when standing on the moon? The masses of
the earth and the moon are respectively,5:98 x 1024kg and 7:36 x 1022kg. Their radii are respectively 6:37 x 103km and 1:74 x 103km

Answer 1

A man weighs 750 N on the surface of the Earth. What would be his weight when standing on the Moon? The masses of the earth and the moon are respectively, $5.98 \times 10^{24} \mathrm{~kg}$ and $7.36 \times 10^{22} \mathrm{~kg}$. Their radii are respectively $6.37 \times 10^{3} \mathrm{~km}$ and $1.74 \times 10^{3} \mathrm{~km}$.

Solution.

We write the Newton's law of universal gravitation:

$$F = G \frac {m _ {1} m _ {2}}{r _ {1 , 2}}$$

where $G$ is the gravitational constant. Particularly, for a man on the Earth surface,

$$F _ {E} = G \frac {m M _ {E}}{R _ {E}},$$

and on the Moon surface:

$$F _ {M} = G \frac {m M _ {M}}{R _ {M}},$$

where $m$ is the mass of a man, $M_{E,M}$ and $R_{E,M}$ are the masses and radii of the Earth and the Moon respectively. Dividing these equations one to another, we obtain

$$F _ {M} = F _ {E} \frac {M _ {M} R _ {E} ^ {2}}{M _ {E} R _ {M} ^ {2}}$$

Putting figures into the final equation, we have:

$$F _ {M} = 7 0 0 N \frac {7 . 3 6 \times 1 0 ^ {2 2} k g (6 . 3 7 \times 1 0 ^ {3} k m) ^ {2}}{5 . 9 8 \times 1 0 ^ {2 4} k g (1 . 7 4 \times 1 0 ^ {3} k m) ^ {2}} = 1 2 3. 7 N.$$

Answer.

$$F _ {M} = 1 2 3. 7 N.$$

http://www.AssignmentExpert.com/