Is the Moon's Gravity Calculation Correct?

[Oops]

Is this calculus correct, or do the decimals points have to be converted ?
Thanks so much , cause I'm stuck


The gravitational constant is 6.673 * 10 to the -11th power.
The radius of the moon is 1737400 meters.
Its mass (which is independent of gravity) is 7.35 * 10 to the 22nd power kg.
Now, given that F=ma and F=GmM/r squared, we can set
a = GM/r squared.
So:
a = (6.67300f * (Math.Pow(10, -11)) * 7.36f * Math.Pow(10, 22))/(1737400f*1737400f)
Running this I get a = 1.62704402190015.
Now, since F (the force of gravity) = ma, then for the same object, F1/F2 = ma1/ma2. This means that the ratio of the forces is equal to the ratio of accelerations since the m cancels out: F1/F2 = a1/a2.
Finally, we take acceleration on the earth, which is roughly 9.8m/s/s. Dividing our calculated value for a by that, we get:
1.62704402190015 / 9.8 = 6.02319
And there it is, our 1:6 ratio.

 

[Oops]

Please does anyone have an answer for me. I just read this in another forum,but I don't know if its correct or not.

 

[chroot]

Looks like this person has some trouble with both arithmetic and physics.

- Warren

 

[Oops]

Chroot, thank you so much for answering - I really appreciate it
could you tell me what's wrong with this calculus ?
It's troubling me a lot

Thnak you - I'm kind of getting desperate :(

 

[Oops]

Oh, it suppose to determine why the gravity of the moon is 1/6 of the earth's

 

[Chronos]

It appears you are trying to calculate tidal forces. In that case, you need to use a higher order derivative.

 

[Oops]

Hi Chronos
Thanks for answering
What do you mean ? What's meant by a higher order derivative ?

 

[Iron Sun 254]

It looks like you're just calculating what the acceleration due to gravity is on the surface of the Moon and then comparing it to the on the surface of the Earth. Looks good to me, but what do I know.

 

[chroot]

That's all he was trying to do? Wow, that was ugly for something so simple.

http://www.google.com/search?hl=en&...rth)+/+(radius+of+earth)^2&btnG=Google+Search
http://www.google.com/search?num=30...he+moon)+/+(radius+of+the+moon)^2&btnG=Search

Or, if you want the ratio:

http://www.google.com/search?num=30...e+moon)+/+(radius+of+the+moon)^2)&btnG=Search

- Warren

 

[Oops]

Now I get it. thanks you guys