Question

The mass of moon is 100 time less than the mass of earth and the radius of moon is 4 time less than the radius of earth show that the value of g is 6 time less than the value of g on earth?

Answer 1

Apply the following steps to approximate the ratio between the gravitational field strength on the earth and on the moon.:

• Find an expression for the gravitational field strength
  `g` on the surface of a sphere.
• Find an expression for the ratio between the gravitational field strength on the earth and on the moon, in terms of mass and radius.
• Substitute in the given ratios between the mass and radius of the earth and the moon to find the value of the ratio in
  `g`.

Step-by-step explanation:

The gravitational field strength
`g` around a sphere is:

• proportional to the mass
  `M` of the sphere, and
• inversely proportional to the square of the distance from the center of the sphere.

Specifically, at a distance of
`r` from the center of mass of a sphere of mass
`M` (above the surface of the sphere), the strength of the gravitational field of the sphere would be:

`\displaystyle g &= (G\, M)/(r^(2))`.

Let
`M(\text{earth})` and
`r(\text{earth})` denote the mass and the radius of planet earth. Let
`M(\text{moon})` and
`r(\text{moon})` denote the mass and radius of the moon. Assume that the density of both the earth and the moon are uniform, such that the distance between the surface and the center of mass is equal to the radius.

`\displaystyle g(\text{earth}) &= \frac{G\, M(\text{earth})}{(r(\text{earth}))^(2)}`.

`\displaystyle g(\text{moon}) &= \frac{G\, M(\text{moon})}{(r(\text{moon}))^(2)}`.

The goal is to find the ratio between the gravitational field strength on the moon and on the earth:

`\begin{aligned} \frac{g(\text{earth})}{g(\text{moon})} &= \frac{\displaystyle \frac{G\, M(\text{earth})}{(r(\text{earth}))^(2)}}{\displaystyle \frac{G\, M(\text{moon})}{(r(\text{moon}))^(2)}} \\ &= \left(\frac{M(\text{earth})}{M(\text{moon})}\right)\, \left(\frac{r(\text{moon})}{r(\text{earth})}\right)^(2)\end{aligned}`.

Given the ratios
`(M(\text{earth}) / M(\text{moon}))` and and
`(r(\text{moon}) / r(\text{earth}))`, the value of the ratio
`(g(\text{earth}) / g(\text{moon}))` would be:

`\begin{aligned} \frac{g(\text{earth})}{g(\text{moon})} &= \left(\frac{M(\text{earth})}{M(\text{moon})}\right)\, \left(\frac{r(\text{moon})}{r(\text{earth})}\right)^(2) =(100)/(16) \approx 6.25\end{aligned}`.

In other words, the gravitational field strength on the moon would be
`6.25` times that on the earth under the assumptions.