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I need someone good at astronomy

  1. Apr 15, 2014 #1
    I'm not a very scientifically or mathematically advanced person, so try not to bash me for this question, but I needed to know the answer for a certain debate I'm having.

    If a planet or 1 km in diameter had 10 times the gravity of Earth, what would be its mass and/or density?

    P.S. Please label what said object's mass or density would be similar to, if anything.
  2. jcsd
  3. Apr 15, 2014 #2


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    By "10 times the gravity of Earth" you mean that the surface gravity, g, is 10 times stronger than the surface gravity of Earth? In other words, it's g=98m/s^2?

    This is simple to solve for yourself. The surface gravity of a spherical body is given by:


    You can plug in 98m/s^2 for g, and plug in 5km for r, and solve for M (the mass). You have to look up G, the gravitation constant. This is not a homework question is it?
  4. Apr 16, 2014 #3
    Well, yes, I'm sure it may be simple to you, and no, this isn't a homework question, it's more of a debate question. I believe I mean surface gravity as well. I'll try to solve it myself, but I'd appreciate a straightforward answer if you could. (I'm not in college yet.)
  5. Apr 16, 2014 #4


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    I didn't provide a straightforward (number) answer because I'm not allowed to just answer homework questions. Homework questions are supposed to be worked out by the individual, and this question sounds quite like a homework question.

    I'm sorry for my use of the word "simple". I realize sometimes my language is condescending, it's because I've done this for a long time, and these things slip out of my mouth.

    Surface gravity, g, is the acceleration you would feel were you on the surface of this object. On Earth, everything falls to the ground at an accelerating speed, accelerating at a=g=9.8m/s^2 (this means you speed up by 9.8 meters per second every second, in other words, you start at 0 speed, in one second you're moving at 9.8 meters per second, and in two seconds you're moving at 19.6 meters per second, etc).

    Because of the nature of gravity, things, light or heavy (whether a feather or a bowling ball), fall to the ground with the same acceleration (in the absence of air resistance). So, this "surface gravity", g, is a property of the gravitating object itself, and does not care about the object being pulled. This is why I assumed "surface gravity" is what you meant, since you did not mention the objects being pulled.
  6. Apr 16, 2014 #5


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    Even if this is not really homework, it's still a homework-type question. PF rules very strictly prohibit help on this type of question without an honest effort on the part of the person who asks the question, and this is clearly stated in the rules.

    Matterwave has already given you the equation that tells you the answer. No further help can be provided, unless you try something for yourself. If you don't understand every term of that equation, google and wiki will give you good information. This is elementary level math and physics, and if you're debating about this stuff, you should aim to become familiar with this at least. It's like the "teaching a guy to fish vs giving him a fish" idiom - it's better you learn to fish for yourself.
  7. Apr 16, 2014 #6
    Thanks. I wasn't aware of that particular rule. That being said, I'll solve it myself, though, I highly doubt you'd learn these sort of mathematics in elementary school.
  8. Apr 16, 2014 #7


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    Why 5km? I think you mean 0.5km.
    Btw random guy,which grade are you studying in?
  9. Apr 16, 2014 #8


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    If equations are not your cup of tea, you can uderstand the problem more intuitivelly if you realise that:

    1.halving the radius halves the gravity
    2.doubling the density doubles the gravity
    (and vice versa)

    Here, you've got a body that's about 1/13000th the radius of Earth. What is the number you'd have to multiply the density of Earth for the body to have the same gravity as Earth? What number for 10 times higher?

    The relationships described above can be extracted from the equation supplied by Matterwave.
  10. Apr 16, 2014 #9


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    Yea you're right, I think I read his "10 times the gravity..." part and recalled 10km as the diameter instead of 1.
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