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Thread: Gravitation & Celestial Mechanics

  1. #1 Gravitation & Celestial Mechanics 
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    I am confused with some astrophysics stuff (esepically Q1), could anyone help me? I would really appreciate if you do.

    1) Calculate the gravitational potential energy of the moon relative to the earth given that the distance between their centers is 3.94x10^8 m.

    Solution:
    Eg = -G(5.98x10^24)(7.35x10^22) / (3.94x10^8 )
    Eg = -7.40x10^28J

    This is an example that I have seen. But I really really don't understand why Eg will have a negative value. (the moon is ABOVE the earth, right? So shouldn't the value of Eg be POSITIVE?) Where is the reference point of the Eg? (at the center of the earth?) For near earth surface case, Eg=mgh, and Eg of objects above the earth will be positive if you set the reference as ground level.

    2) Calculate the change in gravitational potental energy for a 1.0-kg mass lifted 1.0x10^2 km above the surface of Earth. (Mass of earth=5.98x10^24 kg, Radius of earth=6.38x10^6 m)
    I got an answer of 9.6x10^5 J, but the answer in my textbook is 1.0x10^6J. I am having less confidence on this topic. Did I calculate anything wrong or is there a problem with the provided answer?


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  3. #2 Re: Gravitation & Celestial Mechanics 
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    Quote Originally Posted by kingwinner
    I am confused with some astrophysics stuff (esepically Q1), could anyone help me? I would really appreciate if you do.

    1) Calculate the gravitational potential energy of the moon relative to the earth given that the distance between their centers is 3.94x10^8 m.

    Solution:
    Eg = -G(5.98x10^24)(7.35x10^22) / (3.94x10^8 )
    Eg = -7.40x10^28J

    This is an example that I have seen. But I really really don't understand why Eg will have a negative value. (the moon is ABOVE the earth, right? So shouldn't the value of Eg be POSITIVE?) Where is the reference point of the Eg? (at the center of the earth?) For near earth surface case, Eg=mgh, and Eg of objects above the earth will be positive if you set the reference as ground level.
    Excellent question.

    The first you have to understand is that potential energy is relative. In other words it is only a difference in potential energy that really has any significance. You may be familiar with this formula used on the surface of the earth over short distances for potential energy = m g h, where m is the mass of the object, g is the gravitational acceleration and h is the height. But where do you measure the height from? the answer is from anywhere you like, because it is only the difference from one height to another that counts anyway.

    The second thing you have to understand is that this difference in potential energy, when mass m goes from a distance of r1 from a mass M to a distance of r2, is calculated for larger distances by the following formula:
    difference in Potential energy = G M m (1/r1 - 1/r2).
    So the difficulty arises when we want a formula for the potential energy and not for a difference in potential energy. We want a function f of M, m and r for potential energy so that when we subtact the potential energy of our old distance r1 from the potential energy of our new distance r2 we get this same difference above.
    difference in Potential energy = f(M,m,r2) - f(M,m,r1).
    But the only way to do this is to use f(M,m,r) = - G M m / r.

    The problem is that as the distance r gets shorter the potential energy decreases without limit. So there is no place to assign a zero of potential energy. You cannot make r=0 the place of potential energy equal zero because there is an infinite increase of potential energy from r=0 to any r greater than 0. In the other direction as r becomes very large the differnce of potential energy approaches 0 and the potential energy approaches a constant. When we use f(M,m,r) = - G M m / r this constant is zero. We could if you wanted make this constant be any number we would like, say 1,000,000 joules by using the formula
    f(M,m,r) = 1,000,000 Joules - G M m / r
    But that just means that we would eventually find a situaltion with large enough masses and short enough distances where the potential energy becomes zero and then negative. Anyway scientist did not see any point in adding such an arbitrarily chosen number to the potential energy, so they just work with potential energies that are negative, after all it is only the difference in potential energy that matters anyway.


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  4. #3 Re: Gravitation & Celestial Mechanics 
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    Quote Originally Posted by kingwinner
    2) Calculate the change in gravitational potental energy for a 1.0-kg mass lifted 1.0x10^2 km above the surface of Earth. (Mass of earth=5.98x10^24 kg, Radius of earth=6.38x10^6 m)
    I got an answer of 9.6x10^5 J, but the answer in my textbook is 1.0x10^6J. I am having less confidence on this topic. Did I calculate anything wrong or is there a problem with the provided answer?
    Looks to me like they just rounded the answer up. As a teacher I can tell if students are really doing the work because I see answers like yours rather than the anwer copied from the book.
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  5. #4  
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    Thank you for your reply!

    1) But the question says "Calculate the gravitational potential energy of the moon relative to the earth...", RELATIVE to the earth. Um...does that mean the zero Eg point is set on the earth's surface or the center of the earth?

    And why do we always want to make Eg to be negative, in general? (i.e. Eg=-GMm/r) Why don't keep it positive?
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    Quote Originally Posted by kingwinner
    Thank you for your reply!

    1) But the question says "Calculate the gravitational potential energy of the moon relative to the earth...", RELATIVE to the earth. Um...does that mean the zero Eg point is set on the earth's surface or the center of the earth?
    No, relative to the earth simply means considering only the the moon and the earth, for you would get a different answer using the masses of the moon and the sun.

    Quote Originally Posted by kingwinner
    And why do we always want to make Eg to be negative, in general? (i.e. Eg=-GMm/r) Why don't keep it positive?
    As I explained fully in my first post, there is no way to give the potential energy a positive number without being completely arbitrary. If you just change the formula by getting rid of the negative sign then something on the ground would have more potential energy than something up in the air and that is obviously wrong.
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  7. #6  
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    Thanks for explaining!

    1) So if we use the formula Eg=-GMm/r, the refernce point have to be at infinity because that's how to formual is derived, right? That is, when we use the formula Eg=-GMm/r, the gravitational potential energy becomes absolute, not relative......
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  8. #7  
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    You got it. With this formula the zero of potential energy is at infinity. That is exactly right.
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  9. #8  
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    "[QUOTE=kingwinner;20590]I am confused with some astrophysics stuff (esepically Q1), could anyone help me? I would really appreciate if you do.

    1) Calculate the gravitational potential energy of the moon relative to the earth given that the distance between their centers is 3.94x10^8 m.

    Solution:
    Eg = -G(5.98x10^24)(7.35x10^22) / (3.94x10^8 )
    Eg = -7.40x10^28J

    This is an example that I have seen. But I really really don't understand why Eg will have a negative value. (the moon is ABOVE the earth, right? So shouldn't the value of Eg be POSITIVE?) Where is the reference point of the Eg? (at the center of the earth?) For near earth surface case, Eg=mgh, and Eg of objects above the earth will be positive if you set the reference as ground level. "

    Sorry I am new on this forum. The reply may be formatted wrong . Whenever two objects attract (as always in case of gravitation , till date ) , the potential energy is negative .
    But , see in electrostatics , when you have to calculate potential energy between electron - electron or proton - proton , it is always positive because they are going to repel .

    And you know that F = - dU/dr . So , if U is positive , then F is negative , hence repulsion and vice versa .
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  10. #9  
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    [QUOTE=human_being;475749]"
    Quote Originally Posted by kingwinner View Post
    I am confused with some astrophysics stuff (esepically Q1), could anyone help me? I would really appreciate if you do.

    1) Calculate the gravitational potential energy of the moon relative to the earth given that the distance between their centers is 3.94x10^8 m.

    Solution:
    Eg = -G(5.98x10^24)(7.35x10^22) / (3.94x10^8 )
    Eg = -7.40x10^28J

    This is an example that I have seen. But I really really don't understand why Eg will have a negative value. (the moon is ABOVE the earth, right? So shouldn't the value of Eg be POSITIVE?) Where is the reference point of the Eg? (at the center of the earth?) For near earth surface case, Eg=mgh, and Eg of objects above the earth will be positive if you set the reference as ground level. "

    Sorry I am new on this forum. The reply may be formatted wrong . Whenever two objects attract (as always in case of gravitation , till date ) , the potential energy is negative .
    But , see in electrostatics , when you have to calculate potential energy between electron - electron or proton - proton , it is always positive because they are going to repel .

    And you know that F = - dU/dr . So , if U is positive , then F is negative , hence repulsion and vice versa .
    MODERATOR NOTE : human_being, welcome to the forum. You may not have noticed, but the thread you replied to is more than 7 years old; it is unlikely that any of the original contributors are still around to read your ( good ! ) reply. It is generally advisable to stick to more recent threads; anything that hasn't gotten a post in more than, let's say, 2 months or so should be considered "dead", and resurrecting dead threads ( "necro-threading" ) is something we normally avoid doing.
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