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 and1:74 x 103km
km
200.5 N
123.7 N
550.4 N
1000.0 N

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, $5.98 \times 10^{24}$ kg and $7.36 \times 10^{22}$ kg. Their radii are respectively $6.37 \times 10^{3}$ km and $1.74 \times 10^{3}$ km.

200.5 N

123.7 N

550.4 N

1000.0 N

Solution.

With an application of the universal gravitation law, for a man on the Earth surface one have:

$$F_{E} = G \frac{m M_{E}}{R_{E}^{3}},$$

and, similarly, in case of Moon:

$$F_{M} = G \frac{m M_{M}}{R_{M}^{2}},$$

where $G$ is the gravitational constant, $m$ is the mass of a man, $R$ and $M$ are the radii and masses of the Earth and the Moon. To evaluate weight on the Moon surface, we will find a proportionality coefficient between forces $F_{E}$ and $F_{M}$. Combining previous two equations, we find

$$F_{M} = F_{E} \frac{M_{M} R_{E}^{2}}{M_{E} R_{M}^{2}} = \alpha F_{E}$$

That we calculate dimensionless coefficient $\alpha$ numerically:

$$\alpha = \frac{7.36 \times 10^{22} \, \text{kg} \left(6.37 \times 10^{3} \, \text{km}\right)^{2}}{5.98 \times 10^{24} \, \text{kg} \left(1.74 \times 10^{3} \, \text{km}\right)^{2}} = 0.165,$$

what means that weight on the Moon surface reaches only $16.5\%$ of the weight on the Earth. Now we find corresponding weights of different objects multiplying their weights $F_{E}$ to $\alpha$.

Answer.

$$750.0 \, N \times 0.165 = 123.8 \, N$$ $$200.5 \, N \times 0.165 = 33.1 \, N$$ $$123.7 \, N \times 0.165 = 20.4 \, N$$ $$550.4 \, N \times 0.165 = 90.8 \, N$$ $$1000.0 \, N \times 0.165 = 165 \, N$$

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