26.0k views
5 votes
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?

1 Answer

7 votes

Answer:

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.

User Asfourhundred
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.