# Satellite above the earth

1. Mar 19, 2017

### EthanVandals

1. The problem statement, all variables and given/known data
If the earth has a radius of 6 x 10^6 meters, how high above earth's surface does a geosynchronous satellite have to be, if it has a mass of 100kg and gravity acts on the satellite with a force of 4.23 x 10^8 Newtons? What is its linear speed? (Assume the orbit is circular)

2. Relevant equations
Not sure.

3. The attempt at a solution
After reading through the problem, I realized that there has to be some clarification on some points. I'm fairly certain that the geosynchronous idea means that if somebody stood on the earth, the satellite would remain directly above them. That means that the satellite is rotating at the same speed as the earth (in terms of the circumference of its orbit). However, since the radius is larger (we don't actually know it, that's something I need to solve for), its linear speed must be significantly faster than the speed of the earth. So then, I found some equations on the internet, but it seems that I'd need the values for the mass of the earth as well to solve it. Is there something I'm missing? Thanks!

2. Mar 19, 2017

### PeroK

I would write down all the quantities you think might be important and not worry too much at this stage about which ones you are given and which ones you don't know. To get you started, you have:

$M$ = mass of the Earth
$m$ = mass of satellite
$R$ = radius of orbit (from centre of the Earth) - this is one you want to find
$v$ = speed of orbit - this is the other you want to find

Can you complete the list?

3. Mar 19, 2017

### EthanVandals

The mass of the earth is not given in the problem, the mass of the satellite is 100kg, the radius of the orbit has to be directly proportional to how much centripetal force there is keeping the satellite in place. That force is the 4.23 x 10^8 value. So to find the radius, I would need to find the velocity, which would be proportional to how fast the earth spins...But how can I find exactly how quickly the satellite is moving above the earth without knowing the distance that it is from the earth? That's where I'm getting tripped up. In order to find velocity, I need the radius, and in order to find the radius, I need velocity, based on the equation Centripetal Force = (mv^2)/r. Correct me if I'm wrong of course, I'm not too good at this whole physics thing...

4. Mar 19, 2017

### Staff: Mentor

That force on the satellite seems rather large given the presumed location high above the Earth's surface. 100kg at the Earth's surface would only weigh about 980 N. Check your source.

5. Mar 19, 2017

### PeroK

Let me ask you a straight question. Sorry to be blunt:

Which part of that sentence I've underlined don't you undertstand?

6. Mar 19, 2017

### EthanVandals

I'm just getting those values from the problem that my physics professor gave us. I'm not getting numbers from anywhere else, so that's why I'm somewhat confused.

Okay, here's what I know Perok:
Earth's Radius: 6 x 10^6 meters
Satellite Mass: 100kg
Force due to Gravity: 4.23 x 10^8

Those are all the values I have. I could solve for more things like the circumference of the earth, but would that really be important in this scenario? I need the circumference of the satellite's orbit I believe. And don't worry, being blunt is good. I am so awful at understanding this stuff that it helps me a lot.

7. Mar 19, 2017

### PeroK

Okay, so you've added two more to my list.

$M$ = mass of the Earth
$m$ = mass of satellite
$R$ = radius of orbit (from centre of the Earth) - this is one you want to find
$v$ = speed of orbit - this is the other you want to find
$F_g$ = force due to gravity on the satellite
$R_E$ = radius of the Earth

I'll give you two more and let you add the last one:

$G$ = universal gravitational constant
$F_c$ = centripetal force required to keep satellite in a circular orbit

One to go. It's an important one. It's not stated explicitly in the problem, but you worked in out. Hint: geosynchronous.

8. Mar 19, 2017

### Staff: Mentor

Well, as long as you realize that any answers you get will be completely nonsensical, go for it. Don't be surprised if the required speed ends up being many times the speed of light...

9. Mar 19, 2017

### PeroK

@EthanVandals

It's still worth doing the problem a) by looking up $G$ and $M$ and b) (without looking these up) getting the answer in terms of $F_g$. Then, once you're given the correct value for $F_g$ you are good to go.

Solving these problems is similar regardless of what information you are given and what you need to find.

It's up to you whether you want to continue, but clearly the force is many orders of magnitude too big. So, you are only going to get an algebraic answer at this stage.

10. Mar 19, 2017

### Staff: Mentor

Actually, you don't need to look anything up, including the mass of the Earth or G. It's enough that you are already given the gravitational force and that the satellite is geosynchronous. The answers will still be silly, of course.

11. Mar 19, 2017

### EthanVandals

I think the value of 4.23 x 10^8 is already in the format of Fg, since the professor gave it to us in Newtons instead of m/s^2. You said I was close with the geosynchronus idea, and as far as I can think, the earth rotates 2(pi)r every 86,400 seconds, so the satellite will ALSO have to complete a rotation every 86,400 seconds. So the satellite's angular velocity is (2(pi)r)/86,400 seconds. So putting that into the Fc equation, I get (100)(((2(pi)r)^2)/86,400)/r = 4.23 x 10^8. Solving for r, I get the answer r = 7.791960756.

12. Mar 19, 2017

### PeroK

Hmm! Leaving aside the errors in your calculation, you've already been told that the numerical answer will be absurd, so there is no point in calculating it. The angular velocity doesn't depend on $r$ only on the period.

PS IF you are unable to work algebraically, there is no point in continuing with this problem. As the numbers you have been given are clearly wrong, there is no point in plugging them into any equations.

There is actually a lot of physics to be learned from a problem like this, but not sadly by plugging and chugging!

Last edited: Mar 19, 2017