# Homework Help: Falling at the same rate

1. Feb 3, 2007

### aerogurl2

Hi! This maybe a simple question that shouldn't be asked, but I don't know about it. So the question is "would you fall at the same rate on the Moon as on Earth? Explain"
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. Feb 3, 2007

### Kurdt

Staff Emeritus
You are right to consider gravity. What do you know about gravity and how could you use it to find out how fast an object would fall?

3. Feb 3, 2007

### aerogurl2

gravity is on earth which is 9.8 m/s^2 which effects the force because f = ma. but there is no gravity on the moon. thus everything there is at a free fall. so the objects won't fall at the same rate as on the earth? am i on the right track?

4. Feb 3, 2007

### Kurdt

Staff Emeritus
Well you're kind of headed there. The two equations I was looking for was:

$$F = ma$$ and $$F=G\frac{mM}{r^2}$$

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. Feb 3, 2007

### aerogurl2

i've seen the 2nd equation and know that it has somthing to do witht he universal law of gravitiation, but other then that i don't really know what it means.

6. Feb 3, 2007

### Kurdt

Staff Emeritus
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 Me and a body with mass m is falling toward the Earth. What is the acceleration on that object given that the force between the object and the Earth is given by the universal law of gravity?

Then do the same for the moon with mass Mm. Are the accelerations the same?

7. Feb 3, 2007

### hover

Just to add on what the units are-
$$F=G\frac{mM}{r^2}$$
G = 6.67300 × 10-11
M or m = mass measured in kilograms
R = Distance from center in meters

8. Feb 3, 2007

### Kurdt

Staff Emeritus
If you want to state what the units are then G has units of m3 kg-1 s-2

9. Feb 3, 2007

### aerogurl2

?? ma = GmM/r^2 so G, m and r are the same and only M is different becuase of planet's mass. so M = a. then a is greater on earth becuase it's mass is greater then the moon. thus acceleration is different so rate of object is different .??

10. Feb 3, 2007

### hover

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

Last edited: Feb 3, 2007
11. Feb 3, 2007

### hover

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-
$$F=G\frac{M}{r^2}$$

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

12. Feb 3, 2007

### Kurdt

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

$$a=G \frac{M}{r^2}$$

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. Feb 3, 2007

### Kurdt

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

14. Feb 3, 2007

### hover

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

15. Feb 3, 2007

### hover

Just for an example of the equation.
$$a=G\frac{M}{r^2}$$

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
$$a=6.67*10-11\frac{5.9742*10^24}{6378100^2}$$

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. Feb 3, 2007

### mace2

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

17. Feb 4, 2007

### Kurdt

Staff Emeritus
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.