Question

A new planet MPSM, Mystery Planet of Spartan Men, is discovered following the orbital path of the Mystery Planet of Amazon Women. The gravitational field of the MPSM is measured during a satellite fly-by as 1.33 m/s2 at a distance of 5 planetary radii from the center of MPSM. What is the expected strength of the planet's gravity on its surface?

Answer 1

`33.25 m/s^2`

Step-by-step explanation:

The strength of the gravitational field at a certain distance
`r` from the centre of a planet is given by:

`g=(GM)/(r^2)`

where:

G is the gravitational constant

M is the mass of the planet

The equation can be rewritten as follows:

`gr^2 = GM`

The term on the right is a constant term, so we can write:

`g_1 r_1^2 = g_2 r_2^2`

where

`g_1` is the gravitational field strength at distance
`r_1`

`g_2` is the gravitational field strength at distance
`r_2`

Here we have:

When
`r_1 = 5R` (the distance is 5 planetary radii), the gravitational field strength is

`g_1 = 1.33 m/s^2`

where R is the radius of the planet.

Instead at the surface

`r_2 =R`

And so the strength of the gravitational field at the surface is:

`g_2 = (g_1 r_1^2)/(r_2^2)=((1.33)(5R)^2)/(R^2)=(1.33)(25)=33.25 m/s^2`