# Gravity Problem - satellite

1. Oct 30, 2008

### maniacp08

Gravity Problem -- satellite

A satellite with a mass of 250 kg moves in a circular orbit 8.00 x 10^7 m above the Earth's surface.
(a) What is the gravitational force on the satellite?
N
(b) What is the speed of the satellite?
km/s
(c) What is the period of the satellite?
h

Relevant Equations
gravitational Force = G*m1*m2 / r^2
G = 6.67 x 10^-11 Nm^2/kg^2

The gravitational force is given but Im only given the m1 of the satellite, is m2 the mass of Earth 5.9742 × 10^24 kilograms?
How would I derive the radius from the orbit?

for Part b, I would need the answer for part a right?

for part c:
Would I use this formula to calculate the period?
T^2 = (4pi^2 / G*m) * r^3

Last edited: Oct 30, 2008
2. Oct 30, 2008

### cepheid

Staff Emeritus
Re: Gravity Problem -- satellite

For part a, you're given the altitude of the orbit, which is pretty much the same as the radius, except that it's measured from the surface of the earth, not from its centre. SO, you have to add the radius of the earth to this altitude.

As far as the mass of the earth goes...did you get it from a reliable source?

Yes, for part b you need the answer to part a, if you are going to use the fact that gravity is what provides the required centripetal force here to answer the question.

I don't know where that formula for the period comes from, and I'm too lazy to check its dervation. I'd take a much simpler approach. You know the circumference of the orbit (ie the distance covered), and you know the speed of the orbit (from part b). time = distance/speed.

3. Oct 30, 2008

### maniacp08

Re: Gravity Problem -- satellite

Ok, I got the mass of the earth 5.98 × 10^24 kilograms from the textbook.

So the raidus of the Earth is 6.37 x 10^6m which my textbook says.
So I add 8.00 x 10^7 m to it? So is 86,370,000 which is 86.37 x 10^6

gravitational Force = G*m1*m2 / r^2
G = 6.67 x 10^-11 Nm^2/kg^2

F = (6.67 x 10^-11)(5.98 × 10^24)( 250) / (86.37 x 10^6)^2
= (39.886610x^13)( 250) / (7459.7769 x 10^12)
= 9971.65 x 10^13 / (7459.7769 x 10^12)
= 1.336722282 x 10 = 13.36722282 approx = 13.37N

13.37N is the force of gravity
to find the speed, I can do F = MA?
13.37N = m * v^2/r

What would r be in this case?

4. Oct 30, 2008

### cepheid

Staff Emeritus
Re: Gravity Problem -- satellite

Why so hesitant? Does my explanation make sense to you? Please don't just blindly do what I say. Make sure you understand it. Draw a picture. If the satellite is 80 000 km above the *surface* of the earth as stated, then how far is it from the *centre?* What additional piece of info do you need to know to answer that? Now does it make sense?

Ummm...yeah. Basically you can do what you did, which is to equate the graviational force to the centripetal force. The reason for doing so is that gravity is what is providing the centripetal force required for circular motion in this case.

This should really be obvious if you think back to the context in which centripetal force was introduced. It was introduced in the context of circular motion. If something is moving in a circle, it must have a centripetal force acting on it. So what 's r, then?

5. Oct 30, 2008

### maniacp08

Re: Gravity Problem -- satellite

Yes, it makes sense, thanks for explaining.
And I got what r is now that you related it as a circle.

For part C:
What you said earlier
time = distance/speed
where distance is the orbits circumference/distance covered
It would be 2Pi*(86.37 x 10^6) / 2.15km
I get a big answer of 252528.6976s, I divide by 60 since the answer is Hr.
and I get 4208.811. What did I do wrong?

6. Oct 30, 2008

### cepheid

Staff Emeritus
Re: Gravity Problem -- satellite

what do you mean by "distance is the orbits circumference/distance covered?"

The distance in travelled in one orbit is the circumference of the orbit. That's it. So where does the 2.15 km come from???

Again, time=distance/speed, but I don't see a speed in your calculation. What was your speed from part b?

7. Oct 30, 2008

### maniacp08

Re: Gravity Problem -- satellite

I found the speed from part b 2.15 km/s

circumference of the orbit is 2Pi*(86.37 x 10^6)

so time = 2Pi*(86.37 x 10^6) / 2.15 km/s

8. Oct 30, 2008

### cepheid

Staff Emeritus
Re: Gravity Problem -- satellite

I don't know if there's anything wrong with your method. But each part depends upon the previous one, so you'd better check your math.