Finding the net gravitational force on a rocket....

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To find the net gravitational force on a rocket located between the Earth and the moon, the masses of the Earth (5.98 x 10^24 kg) and the moon (7.35 x 10^22 kg) are used with the gravitational force formula Fnet = Gm1m2/r^2. The rocket, with a mass of 1200 kg, is 3.0 x 10^8 m from the Earth's center, placing it 8.4 x 10^7 m from the moon's center. The gravitational forces from both celestial bodies act in opposite directions, necessitating a subtraction of their magnitudes to determine the net force. The stronger gravitational pull from the Earth confirms that the rocket is indeed closer to the Earth than to the moon. The calculations ultimately clarify the distances involved and the net gravitational force acting on the rocket.
Specter
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Homework Statement


The Earth has a mass of 5.98 x 1024 kg and the moon has a mass of 7.35 x 1022 kg. The distance from the centre of the moon to the centre of the Earth is 3.84 x 108 m. A rocket with a total mass of 1200 kg is 3.0 x 108 m from the centre of the Earth and directly in between the Earth and the moon. Find the net gravitational force on the rocket from the Earth and the moon.

Homework Equations


Fnet = Gm1m2/r2

The Attempt at a Solution


m1 (earth)=5.98 x 1024 kg
m2 (moon)=7.35 x 1022 kg
m3 (rocket)=1200kg

m1 is 3.85 x 108m from the centre of m2
m3 is 3.0 x 108 m from the centre of m1

My problem is trying to figure out how far the rocket(m3) is from the moon(m2). The question says that the rocket is directly in between the Earth and the moon, so does that mean that the distance is 3.0 x 108 m to the centre of the moon and to the centre of the earth?

Next I would use Fnet = Gm1m3/r13 + Gm2m3/r23 and solve the question. In the example I read it said "add their magnitudes since both forces act in the same direction". In this question the forces would be acting in different directions, one toward the moon and the other toward earth. Does this mean I subtract the magnitudes instead of adding?
 
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Specter said:
The question says that the rocket is directly in between the Earth and the moon, so does that mean that the distance is 3.0 x 108 m to the centre of the moon and to the centre of the earth?
It is saying that the rocket is somewhere on the line connecting the two centers. Not necessarily at the midpoint of that line.
Specter said:
Does this mean I subtract the magnitudes instead of adding?
That will give you the magnitude of the result. If you want to know its direction, you will have to decide which pull is stronger.
 
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jbriggs444 said:
It is saying that the rocket is somewhere on the line connecting the two centers. Not necessarily at the midpoint of that line.

That will give you the magnitude of the result. If you want to know its direction, you will have to decide which pull is stronger.

The pull of the Earth would be stronger. So if the Earth's pull is stronger than the moons the rocket would be closer to the Earth than the moon, right? But how can I figure out the distance knowing this?
 
Specter said:
So if the Earth's pull is stronger than the moons the rocket would be closer to the Earth than the moon, right?
The position of the rocket is what it is regardless of the force it is subject to.
 
jbriggs444 said:
The position of the rocket is what it is regardless of the force it is subject to.
I think I figured it out. I was making it way harder then it needed to be...

3.84 x 108-3.0 x 108=8.4 x 107
 
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Specter said:
I think I figured it out. I was making it way harder then it needed to be...

3.84 x 108-3.0 x 108=8.4 x 107
Yep. That's the distance of the rocket from the center of the moon.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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