- #1

aerogurl2

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I think that the rate would be the same on moon and earth. i was thinking it had something to do with the universal law of gravitation, but i think i am off track. can someone please help me?

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- Thread starter aerogurl2
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- #1

aerogurl2

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I think that the rate would be the same on moon and earth. i was thinking it had something to do with the universal law of gravitation, but i think i am off track. can someone please help me?

- #2

Kurdt

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- #3

aerogurl2

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- #4

Kurdt

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[tex] F = ma [/tex] and [tex] F=G\frac{mM}{r^2} [/tex]

It is not true to say that the moon has no gravity. The moon exerts a gravitational force on an object because it has mass like the Earth.

Have you seen the second equation before?

- #5

aerogurl2

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- #6

Kurdt

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Ok. The second equation is the universal law of gravitation. It is the force between two objects one of mass m and another of mass M separated by distance r. G is just a constant. So if this second equation is the force between two objects consider the following.

Say the Earth was mass M

Then do the same for the moon with mass M

- #7

hover

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Ok. The second equation is the universal law of gravitation. It is the force between two objects one of mass m and another of mass M separated by distance r. G is just a constant. So if this second equation is the force between two objects consider the following. ?

Just to add on what the units are-

[tex] F=G\frac{mM}{r^2} [/tex]

G = 6.67300 × 10-11

M or m = mass measured in kilograms

R = Distance from center in meters

- #8

Kurdt

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Just to add on what the units are-

[tex] F=G\frac{mM}{r^2} [/tex]

G = 6.67300 × 10-11

M or m = mass measured in kilograms

R = Distance from center in meters

If you want to state what the units are then G has units of m

- #9

aerogurl2

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- #10

hover

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If you want to state what the units are then G has units of m^{3}kg^{-1}s^{-2}

woops, forgot that Also the outcome force is measured in Newtons.

Last edited:

- #11

hover

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M = a? Your saying mass = acceleration?? Thats not correct.

You are right that acceleration is greater here on Earth than the moon, though.

If you really want to know acceleration caused by a mass then use the equation-

[tex] F=G\frac{M}{r^2} [/tex]

units are the same but the outcome is m/s instead of Newtons.

- #12

Kurdt

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You are almost correct. With the highlighted equation you can easily obtain acceleration by cancelling the mass of the falling object to give.

[tex] a=G \frac{M}{r^2} [/tex]

As you correctly deduced the rate of falling on the moon and Earth will therefore be different because the masses of the Earth and Moon are different.

- #13

Kurdt

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If you really want to know acceleration caused by a mass then use the equation-

[tex] F=G\frac{M}{r^2} [/tex]

That is the acceleration due to gravity not the force as you have implied with your notation.

- #14

hover

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That is the acceleration due to gravity not the force as you have implied with your notation.

What are you talking about?? Of course its due to gravity, i don't know what your implying.

- #15

hover

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[tex] a=G\frac{M}{r^2} [/tex]

The Earth has a mass of 5.9742*10^24 kilograms. The radius of the Earth is 6378100 meters. Of course G is 6.67*10^-11. So plugging that in you have

[tex] a=6.67*10-11\frac{5.9742*10^24}{6378100^2} [/tex]

So here we see that acceleration here is 9.8m/s which is the true acceleration of gravity on earth.

Since i did the Earth try doing the moon.

- #16

mace2

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What are you talking about?? Of course its due to gravity, i don't know what your implying.

He's implying that it should have been a = GM/r^2, not F = GM/r^2.

- #17

Kurdt

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What are you talking about?? Of course its due to gravity, i don't know what your implying.

I was saying your notation implied force was given by that equation which it is not. Its acceleration that is given by that equation just as Mace has replied.

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