# Help with a Universal Gravitational Constant

1. Jan 6, 2016

### Atominate

I am doing an Astronomy GCSE. I have to work out the orbital period for a satellite. I have got quite far with it, but I am really stuck. I have spent hours at it! This is the question:-
A space laboratory is in circular orbit around the Earth at a distance of 6000km from the Earth's centre; its period is 2hr 12min. An orbital telescope is to be placed into orbit at 12,000km. Calculate its expected period.
I am at the point where I need to square root 4(pi)²r³/GM = T
A few questions. Is r the radius of the Earth (from core to surface) or core to satellite? How do I calculate GM? I know that M is the mass of the Earth (5.972 × 10^24 kg) but how do I work out G (the Universal Gravitational Constant?). I really need help and it is very frustrating. Any help or advice would be very much appreciated. Thanks!

2. Jan 6, 2016

"r" is usually said to be the distance from the core of Earth to the satellite. So r should be something like:

$r = R + h$
Where R is the radius of Earth and h is the height of the satellite above Earth's surface.

I'm assuming you have to calculate G yourself, correct? You have sufficient information to find G with the information provided for the space laboratory.

3. Jan 6, 2016

4. Jan 6, 2016

### SteamKing

Staff Emeritus
The universal gravitational constant G is one of those fundamental numbers which describes the universe. You don't calculate it; its value is determined experimentally.

For your satellite problem, you just look up the value of G, either in a book or online.

https://en.wikipedia.org/wiki/Gravitational_constant

5. Jan 6, 2016

### TonyS

It is possible to solve your problem without knowing the value of G or the mass of the Earth.

6. Jan 6, 2016

### Atominate

Okay. That's very helpful. Thank you.

7. Jan 6, 2016

### Atominate

I
I looked it up online, but it looks like M, kg and s are required?

8. Jan 6, 2016

9. Jan 6, 2016

### TonyS

The physics of the problem allows you to find an expression for the orbital period as a function of r, that involves your unknown constants.
You know the values of T and r for one particular case, namely the spacestation. The same relation between T and r also holds for the telescope, with the same bunch of constants, so you can eliminate the constants by dividing T(telescope) by T(spacestation).
(Sorry I could make this clearer if I'd learnt to use latex)

10. Jan 6, 2016

### SteamKing

Staff Emeritus
Well, let's take a look at your formula:

G = 6.67408 × 10-11 m3 / kg-s2

You want to find T, which is measured in seconds - check.
You'll have to look up the mass of the earth in kilograms - check.
You know the distance from the center of the earth to the satellite, but you need this in meters instead of kilometers - check.

Looks like G has all the right units.

From a practical standpoint, TonyS' post gives you a good tip for finding the new orbital period.

OTOH, you need to study up on fundamental constants, like G, if you want to do well in your astronomy course.

11. Jan 6, 2016

### Mister T

Use ratio reasoning. The radius of the orbit is doubled, so what happens to the period?

12. Jan 7, 2016

### Atominate

Is the period doubled? That would make everything a lot simpler! Thanks.

13. Jan 7, 2016

### Atominate

Thanks very much. That helps a lot!
P.S. I completely agree. I am lacking with the more mathematical part of course. I will do further learning!

14. Jan 7, 2016

### Atominate

It is very clear now! Thanks.

15. Jan 7, 2016

### ehild

16. Jan 7, 2016

### TonyS

The square of the period is proportional to the cube of the radius, as you indicated in your original post, aside from a typo, I think. This is one of Kepler's laws, by the way.

17. Jan 7, 2016

### Atominate

Kepler's third law says that T2/R3 = constant = 4π2 /GM. I was okay with that part, it was just that I was unsure about GM, but I'm okay now as everyone has helped a lot with their replies.

18. Jan 7, 2016

### Mister T

Since 2 hours 12 minutes is 12.2 hours, we have

$\frac{T^2}{R^3}=\frac{(12.2 \ \mathrm{h})^2}{(6000 \ \mathrm{km})^3}$.

That is the constant of proportionality. So when $R$ doubles to 12 000 km, $T$ does not double to 24.4 hours.