# Universal Gravitation and Gravitational Field Question

1. Jun 3, 2012

### EE123

1. The problem statement, all variables and given/known data
The earth has a mass of 5.98x10^24 kg and the moon has a mass of 7.35x10^22 kg. The distance from the centre of the moon to the centre of the earth is 3.84x10^8 m. A rocket with a total mass of 1200 kg is 3.0x10^8 m from the centre of the earth and directly in between earth and the moon. Find the net gravitional force on the rocket from the earth and moon.

2. Relevant equations

Fg = Gm1m2 / r^2

3. The attempt at a solution

I think the above is the equation I'm suppose to be using. I have hard time grasping this concept :S.

And this is what I did

Fg on the rocket to the earth:

Fg = (6.67x10^-11)(5.98x10^24)(1200) / (3.0x10^8)^2

Fg = 5.3182 N

Fg on the rocket to the moon:

Fg = (6.67x10^-11)(7.35x10^22)(1200) / (3.0x10^8 + 3.84x10^8)^2

Fg = (6.67x10^-11)(7.35x10^22)(1200) / (6.84 x10^9)^2

Fg = 1.2574 x 10^-4 N

Fgnet = 5.3182 N + 1.2574 x 10^-4 N

Fgnet = 5.3183 N

I'm not sure if this is correct or not :S. From my understanding it seems to be. Please help.

2. Jun 3, 2012

### Staff: Mentor

Looks good.

You want the distance from rocket to moon. How might you figure that out? Draw a diagram!

Why did you add the forces? Do they point in the same direction?

3. Jun 3, 2012

### Sleepy_time

Hi EE123. The gravitational force due to the Earth seems to be correct. But for the moon, the equation $F=\frac{Gm_1m_2}{r^2}$ where r is the separation of mass 1 and mass 2. The rocket is between the Earth and the Moon and you know the distance from the Earth to the rocket and from the Earth to the moon, so you should be able to get r from that, it isn't 3.0x10^8 + 3.84x10^8. Also another thing is the force on the rocket due to the moon is in the opposite direction to the force on the rocket due to the Earth, so the net force won't be adding both values.

4. Jun 3, 2012

### EE123

Hi guys and thank you!

And thank you very much for clarification sleepy_time, I appreciate it.

So would it be this then?

distance from rocket to the moon:

Δd = 3.84E8 - 3.0E 8
Δd = 8.4E7 m

Thus,

Fg = (6.67E-11)(7.35E22)(1200) / (8.4E7)^2

Fg = 0.83375 N

FgNet = 5.3182 + (-0.83375 N)

FgNet = 4.48445 N

FgNet = 4.50 N

Can you please explain the following: "the force on the rocket due to the moon is in the opposite direction to the force on the rocket due to the Earth"

5. Jun 3, 2012

### cepheid

Staff Emeritus

Well, since the rocket lies on a straight line in between Earth and the moon, the moon pulls on the rocket in one direction (towards the moon), and the Earth pulls on the rocket in the exact opposite direction (towards the Earth). This is shown in the diagram below. In the diagram, "E" is the earth, "M" is the moon, "R" is the rocket, and arrows show the directions of the two gravitational forces on the rocket. They oppose each other.

Code (Text):
E <--------- R ------> M

6. Jun 3, 2012

### EE123

Okay, thank you! So is my calculation correct??

7. Jun 3, 2012

### Sleepy_time

Yes I believe it is. Just careful with rounding your answer. Rounding 4.48446.... will give 4.5N (1d.p.) or 4.48N (2d.p.)