[SOLVED] Gravitation & Escape Speed - Stone leaving Earth and reached the Moon.

Icetray

Hello everyone, I have one question on gravitation and escape speed. In the earlier part of the question, the minimum speed for the stone to leave the Earth is calculated and its final destination is the Moon.

In the next part, the question asks if the stone will hit the Moon with a speed greater than or smaller than the minimum speed calculated in the previous part.

Can anyone help me with the above question by providing a suitable explanation as well?

Many thanks in advance.

Irid

Have you tried the conservation of energy law?

Icetray

To explain? Actually I have not. I don't even know if the speed will be greater than or less than the original escape speed that was calculated in the earlier part of the question.

Irid

Well, conservation of energy will enable to calculate the value of the speed, so you'll know if it is more or less smth else. Do you know the formula for gravitational potential energy and also for kinetic energy?

Icetray

I believe that you have misunderstood my question. I have already calculated the escape speed for the first part. I just need the second pat solved which asks if the speed at which the stone will hit the Moon's surface will be greater than or smaller than that of the speed it was launched (ie. the escape speed which I had already calculated earlier).

I apologize for the misunderstanding.

Irid

Oh yes, I see, my bad. Maybe there's a shorter way, but I'd still calculate the velocity at the Moon's surface to obtain a figure, and then just compare the two.

Icetray

Alright, so what you are saying here is that I should calculate the Moon's escape speed and compare that to that of the Earth's? It's significant smaller and I don't see how this helps me tell what speed the stone will land on the Moon. I think you're still misunderstanding the question.

Irid

I suggest you to calculate the speed of stone as it reaches the Moon's surface (assume a head-on collision for a greatest possible speed).

arildno

You have now TWO gravitational potentials; that from the Earth, and that from the Moon

Letting R be the distance the stone has from the Earth centre, r the distance it has from the Moon centre, conservation of mechanical energy means:
[tex]\frac{1}{2}mv^{2}-\frac{Gm_{e}m}{R}-\frac{Gm_{m}m}{r}=C[/tex]
where m is the stone's mass, me the Earth mass, mm the moon mass and G the universal gravitation constant (C being the constant value of the energy).

See if you can use this.

Icetray

Thank you for your reply.

I do not actually need to solve and calculate a value and I am also unable to due to the lack of information provided in the question. I am, however, required to explain it. Can you help me phrase this in words? (making it a little clearer for me to understand as well ;-) )

Rainbow Child

How did you find the minimum speed for the stone? Didn't you consider a particular point during the stone' s path?

Icetray

We were given a gravitational potential graph from the Moon to the Earth and we were provided with the values for the gravitational potential for the Earth and Moon at their surfaces. Since Escape speed is independent of its mass, I was able to calculate it using the Earth's gravitational potential.

Rainbow Child

Yes but as arildno, pointed out you have TWO gravitational potentials.

That means that the stone feels two forces one from the Earth and one from the Moon.

The escape velocity depends from both.

Icetray

Hmm...it seems that I have failed to consider that. I always thought that the stone just required to escape the pull of the Earth. Since it's heading to the moon anyway, how will the Moon's pull affect the escape speed? Wouldn't the stone just be required to escape Earth's field alone?

Rainbow Child

Think aboout the forces that acting on the stone when you throw it away.

The force from the Earth resists or helps the stone to move? What about the force from the Moon?

And futhermore, what happens to the values of the two forces while the stone is moving?

Icetray

But the minimum velocity of 11.2 m/s helps the stone leave the Earth's atmosphere. After it has left the Earth's atmosphere, the stone will automatically head towards Earth and later get dragged in towards the Moon's field (if this is what you're trying to tell me). At this stage, why would I need bother about the Moon's gravitational potential?

I don't see how it affects the minimum speed. I don't want the stone to escape the Moon's pull as well. I want the stone to land on the Moon.

Sorry, but I am really lost right now. ):

arildno

Here, I must disagree with your interpretation, Rainbow Child:
I think you are to USE the escape velocity, as calculated in the first part, and then calculate the impact velocity on the moon.

For simplicity, I'll regard the distance between the Moon and the Earth, D as a constant.
Let Re be the Earth radius, Rm the moon radius, then the conservation of energy tells us:
[tex]\frac{1}{2}mv_{i}^{2}-\frac{Gm_{e}m}{D+R_{e}}-\frac{Gm_{m}m}{R_{m}}=\frac{1}{2}mv_{e}^{2}-\frac{Gm_{e}m}{R_{e}}-\frac{Gm_{m}m}{D+R_{m}}[/tex]
where ve is the calculated escape velocity, and vi the impact velocity you were to find.

Now, rearrange terms and evaluate what terms are greater than others.