# Regarding Gravitation and Gravitational Fields

• nblu
In summary, a satellite is designed to orbit Earth at an altitude that places it in a gravitational field with a strength of 4.5N/kg. To calculate the distance above the surface of Earth, the equation g=(G*Me)/r^2 is used, resulting in a value of 3.0 x 10^3km. To calculate the acceleration of the satellite and its direction, the equation a = v^2/r is used, where v=sqrt(G*Me/r), which is derived by equating centripetal force with gravitational force. The acceleration in free-fall is 9.8m/s/s only near the surface of the earth, and it is necessary to calculate the velocity to determine

## Homework Statement

A satellite is designed to orbit Earth at an altitude above its surface that will place it in a gravitational field with a strength of 4.5N/kg.
a) Calculate the distance above the surface of Earth at which the satellite must orbit.
b) Assuming the orbit is circular, calculate the acceleration of the satellite and its direction.
c) At what speed must the satellite travel in order to maintain this orbit?

## Homework Equations

g=(G*Me)/r^2, where Me is Earth's mass

## The Attempt at a Solution

For a), I've used the above equation, with given g value(4.5) substituted into the equation, and got 3.0 x 10^3km as my final answer but b) and c) is where I have the problem. In order to calculate the acceleration for b), I believe that I need to calculate the velocity(v=sqrt(G*Me/r)) first then substitute into (a = v^2/r). However, if I end up getting an answer for part b), didn't I just do c) as well? Because for c) I need to calculate the velocity as well.. I'd like to know whether or not I was wrong about this before I get in any further.

Hint:
orbital motion is "free-fall." What is the acceleration of free fall?

The way you calculated b) is not wrong, and it will work. However, how did you arrive at your equation for velocity? To derive it, you would start by equating centripetal force with gravitational force. What is the centripetal acceleration? Can you figure it out without first calculating velocity?

acceleration in free-fall is 9.8m/s/s right?

Only near the surface of the earth, because the gravitational force is GMm/r2 = mg = 9.8m on the surface.

nblu said:
acceleration in free-fall is 9.8m/s/s right?

only where g is 9.8 N/kg