Please a question [Archive]

Oops

Aug21-04, 08:24 PM

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

Aug21-04, 09:59 PM

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

Aug21-04, 11:35 PM

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

- Warren

Oops

Aug22-04, 12:17 AM

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

Aug22-04, 12:19 AM

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

Chronos

Aug22-04, 02:13 AM

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

Oops

Aug22-04, 02:40 PM

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

Iron Sun 254

Aug23-04, 03:56 PM

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

Aug23-04, 04:09 PM

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

http://www.google.com/search?hl=en&ie=UTF-8&q=%28G+*+mass+of+earth%29+%2F+%28radius+of+earth%2 9%5E2&btnG=Google+Search
http://www.google.com/search?num=30&hl=en&lr=&ie=UTF-8&safe=off&q=%28G+*+mass+of+the+moon%29+%2F+%28radius+of+the+ moon%29%5E2&btnG=Search

Or, if you want the ratio:

http://www.google.com/search?num=30&hl=en&lr=&ie=UTF-8&safe=off&q=%28%28G+*+mass+of+earth%29+%2F+%28radius+of+eart h%29%5E2%29+%2F+%28%28G+*+mass+of+the+moon%29+%2F+ %28radius+of+the+moon%29%5E2%29&btnG=Search

- Warren

Oops

Aug24-04, 01:04 AM

Now I get it. thanks you guys