Question

a satellite orbits mars at a distance above its surface equal to 3 times the radius of mars. what is the acceleration of gravity experienced by the satellite compared to the acceleration of gravity experienced on the surface of mars?

Answer 1

The gravitational acceleration in this orbit would be
`(1/16)` the value on the surface.

Step-by-step explanation:

This question can be solved by analyzing the equation for the gravitational field strength outside a sphere.

Assume that the mass of Mars is symmetrically distributed with respect to its center, such that the center of mass of the planet would be at its geometric center.

At a distance of
`r` from the center of mass of a sphere (outside the sphere,) the gravitational field strength, or equivalently the acceleration of gravity, would be:

`\displaystyle (G\, M)/(r^(2))`,

Where:

• 
  `G` is the gravitational constant, and
• 
  `M` is the mass of the sphere (mass of Mars.)

Let
`R` denote the radius of Mars. Under the assumptions, distance from the center of mass of the planet would be
`r = R` on the surface of the planet. The gravitational field strength would be:

`\displaystyle (G\, M)/(R^(2))`.

At a distance of
`3\, R` from the surface of the planet, the distance from the center of mass of the planet would become
`r = R + 3\, R = 4\, R`. The gravitational field strength would be:

`\displaystyle (G\, M)/((4\, R)^(2)) = (1)/(16)\, \left((G\, M)/(R^(2))\right)`.

In other words, the acceleration of gravity in this orbit would be
`(1/16)` the value on the surface.