# A 500 kg satellite experiences a gravitational force of....

• LionLieOn
In summary, the conversation discusses finding the period of a 500 kg satellite in circular orbit around the Earth, given a gravitational force of 3000 N. Two methods are presented, one using the equation V=2(Pi)r/T and the other using ω^2=6^3/GM. It is concluded that either method can be used to calculate the period.
LionLieOn

## Homework Statement

A 500 kg satellite experiences a gravitational force of 3000 N, while moving in a circular orbit around the earth.
c) Find the Period of the orbit

## Homework Equations

So found the period using (Please see the attachment to review my work.) but I also found it by using (V=D/T)
V= 2(Pi)r/T

I was wondering if there's any difference?

## The Attempt at a Solution

#### Attachments

• FCFG.jpg
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LionLieOn said:

## Homework Statement

A 500 kg satellite experiences a gravitational force of 3000 N, while moving in a circular orbit around the earth.
c) Find the Period of the orbit

## Homework Equations

So found the period using (Please see the attachment to review my work.) but I also found it by using (V=D/T)
V= 2(Pi)r/T

I was wondering if there's any difference?

## The Attempt at a Solution

Your result is correct, and after getting V from the centripetal force, you can calculate T from V= 2(Pi)r/T.

ehild said:
Your result is correct, and after getting V from the centripetal force, you can calculate T from V= 2(Pi)r/T.

So either 1 is fine?

LionLieOn said:
So either 1 is fine?
Yes, either one. And there is an even simpler method to calculate the period.
You have two equations, one for Fc in terms of ω, ##Fc=mrω^2## and one for Fg ##Fg=G\frac{mM}{r^2}##, : with Fc=Fg=3000 and m=500.
Isolate r from one of them and substitute the expression for r into the other equation. No need to calculate the numerical value of r. You get the simple formula ##ω^2=\frac{6^3}{GM}##

ehild said:
Yes, either one. And there is an even simpler method to calculate the period.
You have two equations, one for Fc in terms of ω, ##Fc=mrω^2## and one for Fg ##Fg=G\frac{mM}{r^2}##, : with Fc=Fg=3000 and m=500.
Isolate r from one of them and substitute the expression for r into the other equation. No need to calculate the numerical value of r. You get the simple formula ##ω^2=\frac{6^3}{GM}##
Ahh! Ok. Thank you so much for your help :)

## 1. How is the gravitational force calculated for a 500 kg satellite?

The gravitational force on a 500 kg satellite is calculated using the formula F = (G * m1 * m2) / r^2, where G is the universal gravitational constant (6.67 x 10^-11 N*m^2/kg^2), m1 is the mass of the satellite (500 kg), m2 is the mass of the planet or object it is orbiting, and r is the distance between the center of the satellite and the center of the planet or object.

## 2. What factors affect the gravitational force on a 500 kg satellite?

The gravitational force on a 500 kg satellite is affected by the mass of the satellite and the mass of the planet or object it is orbiting, as well as the distance between the two objects. The force decreases as the distance between the objects increases, but increases as the mass of either object increases.

## 3. What is the direction of the gravitational force on a 500 kg satellite?

The gravitational force on a 500 kg satellite is always directed towards the center of the planet or object it is orbiting. This is due to the fact that gravity is a conservative force, meaning it always acts towards the center of mass of the objects involved.

## 4. Can the gravitational force on a 500 kg satellite change during its orbit?

Yes, the gravitational force on a 500 kg satellite can change during its orbit. This is because the distance between the satellite and the planet or object it is orbiting is constantly changing as the satellite moves along its orbit. As the distance changes, the gravitational force also changes accordingly.

## 5. How does the gravitational force on a 500 kg satellite affect its orbit?

The gravitational force on a 500 kg satellite is what keeps it in orbit around a planet or object. Without this force, the satellite would continue in a straight line and leave orbit. The strength of the gravitational force determines the speed and shape of the satellite's orbit.

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