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

[Icetray]

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

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]

Have you tried the conservation of energy law?

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]

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?

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]

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.

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:
$$\frac{1}{2}mv^{2}-\frac{Gm_{e}m}{R}-\frac{Gm_{m}m}{r}=C$$
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]

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:
$$\frac{1}{2}mv^{2}-\frac{Gm_{e}m}{R}-\frac{Gm_{m}m}{r}=C$$
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.

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]

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 ;-) )

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

 

[Icetray]

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

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]

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.

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]

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?

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:
$$\frac{1}{2}mv_{i}^{2}-\frac{Gm_{e}m}{D+R_{e}}-\frac{Gm_{m}m}{R_{m}}=\frac{1}{2}mv_{e}^{2}-\frac{Gm_{e}m}{R_{e}}-\frac{Gm_{m}m}{D+R_{m}}$$
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.

 

[Icetray]

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:
$$\frac{1}{2}mv_{i}^{2}-\frac{Gm_{e}m}{D+R_{e}}-\frac{Gm_{m}m}{R_{m}}=\frac{1}{2}mv_{e}^{2}-\frac{Gm_{e}m}{R_{e}}-\frac{Gm_{m}m}{D+R_{m}}$$
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.

Alright, now here I am really lost. Can you phrase what you have written in just words for me? How did you come up with the equation?

Thank you for being so patient.

 

[Rainbow Child]

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:
$$\frac{1}{2}mv_{i}^{2}-\frac{Gm_{e}m}{D+R_{e}}-\frac{Gm_{m}m}{R_{m}}=\frac{1}{2}mv_{e}^{2}-\frac{Gm_{e}m}{R_{e}}-\frac{Gm_{m}m}{D+R_{m}}$$
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.

Ok! It is nice to disagree, if we can help each other and of course Icetray!

The force from the Earth is resisting the stone's movement and at the same time time it decreases since the distance from the center of the Earth increases. On the other hand the force from the Moon helps the stone's movement and is increasing since the distane from the Moon's center is decreasing.

Thus we must first find the point where the two forces are equal.

$$F_1=F_2\Rightarrow\frac{G\,m_e\,m}{x^2}=\frac{G\,m_m\,m}{(d-x)^2}$$
where $$d$$ the distane between Earth's and Moon's centers.

If the stone reaches there with zero velocity the Moon's attraction will bring it to the surface of the Moon!

 

[Icetray]

Ok! It is nice to disagree, if we can help each other and of course Icetray!

The force from the Earth is resisting the stone's movement and at the same time time it decreases since the distance from the center of the Earth increases. On the other hand the force from the Moon helps the stone's movement and is increasing since the distane from the Moon's center is decreasing.

Thus we must first find the point where the two forces are equal.

$$F_1=F_2\Rightarrow\frac{G\,m_e\,m}{x^2}=\frac{G\,m_m\,m}{(d-x)^2}$$
where $$d$$ the distane between Earth's and Moon's centers.

If the stone reaches there with zero velocity the Moon's attraction will bring it to the surface of the Moon!

Ok, now this whole thing is getting really complicated. Can I refer everyone back to my question? Will the stone hit the moon at the speed it left the Earth and hit the surface of the moon with a greater or smaller velocity than that of which it was lost.

 

[Rainbow Child]

Icetray, in your graph there must be a maximum point. That refers to the distance I was describing in my previous post.

Is there a maximum?

 

[arildno]

What ARE you talking about, rainchild??

You are to find out if vi>ve or ve>vi from the expression I gave.

 

[arildno]

Rearranging, we get:
$$\frac{1}{2}mv_{i}^{2}=\frac{1}{2}mv_{e}^{2}+\frac{Gm_{e}m}{D+R_{e}}-\frac{Gm_{e}m}{R_{e}}+\frac{Gm_{m}}{R_{m}m}-\frac{Gm_{m}m}{D+R_{m}}$$

Now, D is much larger than Re and Rm, so we can ignore those terms including D compared to those who have otherwise equal factors.

We therefore have:
$$\frac{1}{2}mv_{i}^{2}\approx\frac{1}{2}mv_{e}^{2}-\frac{Gm_{e}m}{R_{e}}+\frac{Gm_{m}m}{R_{m}}$$

Now, since the ratio m_{e}/R_{e}>m_{m}/R_{m}, it follows that vi<ve.

 

[Rainbow Child]

arildno, let me ask you something, could I?

You are saying that the minimum excape velocity of a stone which is thrown from the surface of the Earth towards the Moon, is $$v=\sqrt{\frac{2\,G\,M_e}{R_e}}=11.2 Km/s$$?

 

[Icetray]

http://img48.imageshack.us/img48/4483/document001pe9.jpg

Above is the question. It's only a two marks question but everyone is making everything so complicated.

 

[Oerg]

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.

Since you were already given a gravitational potential graph from Earth to moon, then the question has already taken into account both potentials exhibited by the moon and the earth.

If that is the case, then we can say that it will hit the moon with a lower velocity because the gravitational field strength of the Earth is stronger than the moon. To explain it to you clearly, think of the moon as an earth. Now we have 2 Earth's. If we were to throw a stone from Earth 1 to Earth 2, the stone will lose velocity while escaping Earth 1, but gain back its velocity as it is pulled towards Earth 2. Thus, the stone will have the same velocity as it left Earth one and as it hits Earth 2. So for a lower mass like the moon, the stone will hit with a lower velocity due to a lower gravitaiotnal field strength exhibited by the moon.

 

[Icetray]

Since you were already given a gravitational potential graph from Earth to moon, then the question has already taken into account both potentials exhibited by the moon and the earth.

If that is the case, then we can say that it will hit the moon with a lower velocity because the gravitational field strength of the Earth is stronger than the moon. To explain it to you clearly, think of the moon as an earth. Now we have 2 Earth's. If we were to throw a stone from Earth 1 to Earth 2, the stone will lose velocity while escaping Earth 1, but gain back its velocity as it is pulled towards Earth 2. Thus, the stone will have the same velocity as it left Earth one and as it hits Earth 2. So for a lower mass like the moon, the stone will hit with a lower velocity due to a lower gravitaiotnal field strength exhibited by the moon.

Thanks! This was the simple answer I was looking for haha. Everyone ended up making everything so complicated in the end.

 

[Rainbow Child]

Icetray can you upload the Fig. 2.1?
"The gravitational potential graph from the Moon to the Earth."

 

[Icetray]

http://img246.imageshack.us/img246/1232/document002xu6.jpg

There you go Rainbow Child. There is also aa part on why the graph isn't symmetrical. Can I just say that it is because gravitational potential does not vary evenly with distance?

 

[shawshank]

There you go Rainbow Child. There is also aa part on why the graph isn't symmetrical. Can I just say that it is because gravitational potential does not vary evenly with distance?

yea, that you can say because it is inversely proportional to the square of the radius.

 

[Icetray]

yea, that you can say because it is inversely proportional to the square of the radius.

Thank you. Your help is much appreciated. Can anyone guide me on how to close this topic? I click on threat tools but it just sends me to the bottom of the page (on Safari) and I tried editing to topic title and put 'Solved' before my topic title but for some reason it changed back or something today.

 

[HallsofIvy]

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.

Is it actually necessary to do all that calculation? You seem to be saying at the first that that has already been done. It seems to me easy to answer both questions given here from general principals. In order to "escape" from Earth's gravity, an object would have to have enough speed to eventually go out to "infinity". Since that is much farther than to the moon, it clearly requires more initial speed than going to the moon. The minimum speed necessary for the rock to read the moon is lower than the minimum speed necessary to "leave the earth" (escape velocity).

Since the rock on the moon has higher potential energy (relative to the earth) than the rock on the earth, in order that total energy be conserved, the rocks kinetic energy must decline. The speed with which the rock hits the moon is lower than the speed with which it leaves the moon.

 

[Rainbow Child]

Happy new year to everybody!

Thanks for the figure Icetray. Now let me expain why your calculation for the first part is wrong (and why arildno was wrong).

As I can see, there is a first question asking for the gravitational field strength at the point P (the maximum I was talking about). What is so special about this point?

Well then the potential is maximum of minimum the total force is zero. So what could someone say. But look

• The total force from the surface of Earth to point P is towards the Earth so until that point the stone decelerates.
• The total force from point P to the surface of the Moon is towards the Moon so the stone accelerates.

Thus in order for the stone to reach the Moon it has only to reach point P with zero velocity, after that point the total force will accerelates it towards the Moon. Writing down the conservation of energy from the surface of the Earth to point P, you will find the desired escape velocity to be $$v_o=11.045 \,Km/sec$$ not $$v_o=11.2 \,Km/sec$$. Excatly as HallsofIvy posted.

Writing now the conservation of energy from point P to the surface of the Moon, you will find that the final velocity is $$v=2.28\,Km/sec$$