Mass of the moon

1. Jan 26, 2014

calculator20

I'm looking at a textbook problem which has an answer but no working out. I've read the chapter several times and looked at the equations but can't see where to start with it? Can you help?

Assuming the Earth and Moon to be isolated from all other masses, use the following data to estimate the mass of the Moon.

Mass of the Earth = 6.0 x 10^24 kg

Distance between centre of Earth and centre of Moon = 3.8 x 10^8 m

Distance from centre of Earth at which gravitational field is zero = 2.8 x 10^8 m

Answer = 7.4 x 10^22 kg

I think it's the 2 distances throwing me, Why do I need to know when gravity is zero? Do I need to find force? Help! Many thanks.

2. Jan 26, 2014

BvU

Yes, force is the key here. No need to calculate: that's already been done for this peculiar intermediate point. The net force is zero. Earth and moon pull equally hard on any test mass at this point. What you want is a relevant expression (oh, this nice template we all should use!) for these forces in terms like the mass of the earth, the moon, some distances, perhaps a constant, and so on. Then sort out what you know and don't know c.q. want to know. Then it becomes mathematics.

3. Jan 26, 2014

calculator20

Thanks for that, still not quite getting though.

If net force is zero do I equate the 2 versions of GM/r^2? G's cancel giving:

Me/R^2 = Mm/(R-r)^2

6x10^24/(2.8x10^8)^2 x (1x10^8)^2= 7.7x10^23

Close but not right, where am I going wrong?

4. Jan 26, 2014

BvU

Subtle difference. Too big to ignore as you conclude correctly. What else could be involved ?

5. Jan 26, 2014

calculator20

I'm sure I'm starring at it but I have no idea!

6. Jan 26, 2014

BvU

'Kidding'... 7.4 x 10^22 kg and 7.7x10^22 are close. 7.7x10^23 is an order of magnitude too big.
Not subtle at all (for subtle I thought about rotation effects, but forget about that).

Could there be a typo somewhere in the exercise?
As you found mass ratio = distance^-2 ratio. Ratio should be around 80, so distance ratio around 9. Here it is around 3 causing trouble. I'm puzzled.

7. Jan 26, 2014

calculator20

Maybe it is a printing typo. I've double checked the original and I've used the right numbers.

If my method and answer should be the correct ones that's fine. If I am missing something please let me know.

8. Jan 26, 2014

BvU

back to the rotation issue. Google L1 point earth moon L1 to realize that its effect is bigger than I thought...

9. Jan 26, 2014

D H

Staff Emeritus
That's a typo in your textbook. It looks like the authors lost a factor of ten somewhere along the way in creating this problem. That's the location at which the gravitational forces would cancel if the Moon was ten times as massive as it is. If you use a distance of 3.42 x 10^8 m instead of that erroneous value you will get the correct mass.

Last edited: Jan 26, 2014
10. Jan 26, 2014

calculator20

Thank you both, much appreciated!