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

[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