Gravitational satellite question.

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Homework Help Overview

The discussion revolves around a problem involving gravitational attraction, specifically a rock falling towards Earth from a significant distance. The participants explore concepts related to gravitational potential energy and kinetic energy, as well as the implications of the rock's initial state of rest relative to Earth.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation of energy principle and its application to the problem. There are questions about the correct interpretation of the rock's initial conditions, including whether it is in orbit or at rest. Some participants express uncertainty about the initial distance of the rock and its implications for the calculations.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's assumptions. Some guidance has been offered regarding the interpretation of potential energy and the initial conditions of the rock. There is recognition of the need for clarity on the initial distance, with varying interpretations being explored.

Contextual Notes

There is a lack of specific information regarding the initial distance of the rock, which is critical for solving the problem. Participants note that the phrase "far out in space" can lead to different interpretations, affecting the approach to the problem.

al_201314
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Hi guys I need help with this after quite some time

A piece of rock far out in space is relative to the earth. Under the influence of the Earth's gravitational attraction, it begins to fall towards the Earth along a straight adial line. With what speed will the rock hit the earth? (Given radius of Earth = 6.4 x 10^6m)

I couldn't find the radius of orbit of the rock. I suppose since it is at rest relative to Earth it must have the same angular speed? How do I find the radius from here? I figured that its period is not the same as the Earth's as the T^2 = k(r^3).

Thanks a lot!
 
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Falling in straight radial line means that the rock will fall straight down ie. it will not orbit the earth.

Think about the conservation of energy principle.
 
The total energy the rock has is -GMm/r^2 during the orbit where r is radius of orbit am I right? So when it drops straight all these energy will be converted to KE of the rock? So -GMm/r^2 - GMm/(6.4 x 10^6)^2 = 1/2mv^2? However I still can't solve from here as mass of Earth is not given as well as r?

Thanks!
 
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Well although the formula for potential energy you have is correct, you must confirm this as the rock is not orbiting the earth. (You stated that using the formula would be correct as the rock is orbiting the earth, but this is not the case)

Mass of Earth can be assumed known (Google for the precise numerical value). For r: The rock starts out falling from really, really far away. How would you therefore approximate its potential at the beginning of the drop?
 
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Since it is stationary relative to the earth, wouldn't it be orbiting just like a geostationary satellite in orbit?

The potential is 0 at infinity?

Anyway the given answer is 1.1 x 10^4 ms^-1
 
al_201314 said:
Since it is stationary relative to the earth, wouldn't it be orbiting just like a geostationary satellite in orbit?
No. Looking from an inertial earth-centered frame, the rock does not have any velocity at the beginning (you earlier, correctly, assumed that the kinetic energy of the rock was 0). This would not be the case if the rock were orbiting (as it would mean that anglular momentum L = 0, which would in turn mean that the rock weren't orbiting).

The potential is 0 at infinity?
Yes.

Anyway the given answer is 1.1 x 10^4 ms^-1
True. And you should get it from the energy conservation equation you had there a few posts ago.
 
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al_201314 said:
The total energy the rock has is -GMm/r^2 during the orbit where r is radius of orbit am I right?
...

The formula for the potential energy which you were using is not correct.
Go it over before you go on in your calculations.
 
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WigneRacah said:
The formula for the potential energy which you were using is not correct.
Go it over before you go on in your calculations.
I cannot believe that went totally unnoticed to me. Worse than that, I called it correct. Thank you!
 
Unless I'm completely mistaken, this question is unsolvable. The initial distance of the rock is not given. We don't know if by "far out in space" they mean 10 thousand miles or 10 million miles.


Now, there seems to be some suggestion that "at rest relative to the Earth" is to be interpreted as "in geostationary orbit". I believe this is an erroneous interpretation. I believe "at rest relative to the Earth" simply means "no velocity towards (or away from) Earth".
 
  • #10
DaveC426913 said:
Unless I'm completely mistaken, this question is unsolvable. The initial distance of the rock is not given. We don't know if by "far out in space" they mean 10 thousand miles or 10 million miles.
Usually such phrase is interpreted (modeled, approximated) as infinitely far, which applied to this particular problem does give the wanted 11 km/s (escape velocity).
 
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  • #11
Ahh thanks guys I got the answer in just one step. It is basically just the equation for the escape velocity. I guess all my wrong interpretations lead me to think otherwise and I can't believe at my level I got the GPE equation wrong!

Thanks anyway!
 
  • #12
Allow me to ask another one I couldn't really get the answer as well.

Earth has a radius R and value of acceleration of free fall at its surface is g. A planet in another system has same average density as Earth but free fall at its surface is 3g/2. Whats the radius of this planet?

g = Gm/R^2 for earth.

g' = GM/x^2.

They didn't mention anything about the mass how can I go about it? Equating this 2 gave the wrong answer of (2^1/2)R/3.

EDIT: typo.
 
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  • #13
A planet in another system has same average density as Earth but free fall at its surface _________ Whats the radius of this planet?
Are you missing a word or two there?

Start by relating the mass to something given in the problem (density).
 
  • #14
Sorry for the mistake.

And I just managed to solve it. Sorry for the trouble and your tip on the mass helped.

Thanks!
 

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