# Gravity Problem

1. Apr 5, 2008

A 619 kg satellite is in circular orbit 7.84x106 m above the surface of the Earth. Find:

a) the acceleration due to gravity created by the Earth at the distance of the satellite.
correct check mark m/s2

b) the weight of a 70.4 kg astronaut inside the satellite.

Equations:
Fg = G M earth M / r$$^{2}$$

Part A is solved. However, part b is simple enough but i seem to be having trouble with it.

My initial attempt was to use the first part of the answer and use Newton's formula F = ma.

However, my next attempt is to use:

Fg = G M(earth) M(astronaut) / r$$^{2}$$

However, one thing gives me trouble. Should the astronaut's mass be used or should the astronaut and the satellite. If I'm off track I would appreciate someone's help. Thanks.

2. Apr 6, 2008

### Dr. Jekyll

You don't need the weight of the satellite so you should use the astronaut's mass.

There is also the centrifugal force on the astronaut but you probably don't need it.

3. Apr 6, 2008

### HallsofIvy

Staff Emeritus
The "weight" of the astronaut is simply the gravitational force on him at that height.

4. Apr 6, 2008

I have another question relating to this problem

1.the work that was done to put the satellite into this orbit. Assume that it starts at rest on the surface of the earth.

Friction is to be ignored.

I have the orbital speed. Could the work-energy theorem be used?

If so then it would be

W = 1/2 m(satellite) v(orbital)^2 - 1/2 m (0)
W = 1/2 m (satellite) v(orbital)^2

Ive tried using W = F change in distance
where the force is the gravitational force and the distance is the height above the earth's surface but I got the answer wrong. Perhaps I'm missing something.

5. Apr 6, 2008

### D H

Staff Emeritus
You've missed the fact that the Earth is rotating and that the potential energy on the ground and differs from that on orbit.

6. Apr 6, 2008

Right.

So I would use this formula:

$$\Delta$$ E = Ef - Ei

So I would get this:

Ef =

1/2 $$M_{satellite}$$ $$V_{orbital speed}$$$$^{2}$$ - G$$M_{earth}M_{satellite}$$/R

The radius would be the radius of the earth plus the height above the surface

Ei = 1/2 $$M_{satellite}$$ $$V_{initial}$$$$^{2}$$ - G$$M_{earth}M_{satellite}$$/R

The radius would be the radius of the earth. And the initial velocity would be zero since the satellite is at rest.

Ef - Ei = 1/2 $$M_{satellite}$$ $$V_{orbital speed}$$$$^{2}$$ - G$$M_{earth}M_{satellite}$$/R - (- G$$M_{earth}M_{satellite}$$/R)

Last edited: Apr 6, 2008
7. Apr 6, 2008

### D H

Staff Emeritus
The satellite is not at rest initially. The Earth is rotating.

8. Apr 6, 2008

### Gear300

Since the acceleration of the Earth's gravitational force differs at different radial distances from the planet, you need a certain amount of Kinetic Energy to get 7.84x106 m above the surface of the Earth. So the change in kinetic energy of the rocket is equal to negative the change in potential energy...and the work done by the rocket is equal to the change in kinetic energy.