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Mass of a planet using a satellite

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A satellite circles planet X once every 99 min at a mean altitude of 530km. Calculate the mass of the Planet, if its radius is 5.78 * 10^6

    So this is the Question i am having trouble with, these gravity questions are really killing me.


    2. Relevant equations


    4 pi2R2/ T2 = GM/R





    3. The attempt at a solution

    So i changed that equation in the following

    4 pi2R3/ G T2 = M

    and then i change the 99 minuets in to seconds and then i change the 530Km in to M . Now this is where a friend told me that i need to at 530 000m in the the given radius of 5.78 * 10^6. I couldnt understand why he said that though. So going on with what i was told i plugged everything in to the equation and i got

    5.246804357 * 1017Kg as the mass of the planet.

    Now i have doubts about this because i didn't quite know what to do with the 530 000m. I hope someone can correct my mistakes and show me the right solution. These gravitational questions are really giving me a tough time today.

    thanks.
     
  2. jcsd
  3. Oct 18, 2009 #2

    cepheid

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    If you don't understand that, then you don't understand the problem, and you are basically just manipulating formulas without understanding what those formulas mean. That would be very unfortunate. Physics is not about blindly applying formulas to calculate things. Physics is about understanding concepts and expressing them mathematically. Let me ask you something cruisx: Do you *understand* the main concept behind this problem? Do you know why the equality 4π2r2 / T2 = GM/R is true in this situation?

    I'm not trying to pick on you. I'm only asking because, if you don't understand the actual physics, then I am willing to do my best to try and explain.
     
  4. Oct 18, 2009 #3

    No, cepheid i am not fully understanding the actual physics behind that formula, i am have a bit of a hard time with Universal gravitation. Could you please explain to me if you have time?
     
  5. Oct 18, 2009 #4

    cepheid

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    Newton's law of gravitation states that if you have two masses m and M, separated by a distance r, then the strength of the gravitational force between them is given by the relationship:

    [tex] F = \frac{GmM}{r^2} [/tex] ​

    where r is the distance between the centres of the two masses. This is an important point that will come up later.

    Now, if you have an object moving on a circular path, you can show (and I won't give the derivation here -- it is probably in your book) using entirely geometric arguments that that object must have an acceleration that always (at any instant) points towards the centre of the circular path. (That's what 'centripetal' means -- pointing towards the centre). You can also show that if the object moves with a constant speed v and its circular path has radius r, then the magnitude of this centripetal acceleration is given by a = v2/r. If the object has mass m, then Newton's 2nd law says that F = ma, and therefore the magnitude of the centripetal force is F = mv2/r. Now I want to point out that this is a kinematic requirement -- i.e. a centripetal force of this magnitude is required for any sort of uniform circular motion, but the equation doesn't tell you anything about what is causing that centripetal force. It could be because the object is tied to a string, or maybe it's under the influence of gravity, or maybe it is charged and experiencing an electrostatic force, or maybe there's friction, or...

    In our situation, gravity is what is providing the centripetal force. The satellite moves in a circular path because the planet's gravity is constantly pulling on the satellite. This pull is in a direction towards the centre of the planet (and therefore towards the centre of the circular orbit). It is an example of a centripetal force. Now, if the satellite has mass m, then the centripetal force must have magnitude mv2/r. However, since the force is also the gravitational force of attraction between the planet and satellite, then Newton's Law of Gravitation says that its magnitude is given by GmM/r2. Since the centripetal force IS gravity in this situation, these magnitudes must be equal:

    [tex] F_{\textrm{cent}} = F_{\textrm{grav}} [/tex]

    [tex] \frac{mv^2}{r} = \frac{GmM}{r^2} [/tex] ​

    The mass of the satellite, m, cancels from both sides of the equation (which means that the properties of the orbit don't depend upon how big the satellite is!). We are left with:

    [tex] \frac{v^2}{r} = \frac{GM}{r^2} [/tex] ​

    Multiplying both sides of the equation by r, we get:

    [tex] v^2 = \frac{GM}{r} [/tex] ​

    How can we figure out the speed, v, of the satellite? We know how much time the satellite takes to go around in its orbit (which is the orbital period, T). This is given in the problem. Speed = distance/time. So if we can just figure out how much distance is covered in one orbit, we can calculate the speed by dividing that distance by T. Well, the orbit is a circle of radius r. Therefore, from basic geometry, the distance around the orbit is given by 2πr. Speed = distance/time, which means that v =2πr/T. Plugging this expression for the velocity into the previous equation, the result is:

    [tex] \left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r} [/tex] ​

    [tex] \frac{4\pi^2 r^2}{T^2}= \frac{GM}{r} [/tex] ​

    THAT's why this equation is applicable to the problem. Now we just have to solve for M, the planet's mass. We need to calculate r, the orbital radius. Here comes the important point. Remember that in Newton's law of gravitation, the distance r, between the two masses, is the distance between their centres. Therefore, r is the distance between the satellite and the centre of the planet. THAT is why r is given by:

    r = radius of planet + height of orbit.

    I hope that that clears things up.
     
    Last edited: Oct 18, 2009
  6. Oct 18, 2009 #5

    Yes that makes much more sense, thanks for the explanation. So you are saying that r is the distance between their centers so something liek this?

    http://img196.imageshack.us/img196/709/gravl.jpg [Broken]

    Thanks, i am going to check over my solution one more time and see if i can come up with the correct answer.
     
    Last edited by a moderator: May 4, 2017
  7. Oct 18, 2009 #6

    cepheid

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    [/URL]

    Yeah.

    Well, good luck. Once you have the concepts down, it's just algebra from there.
     
    Last edited by a moderator: May 4, 2017
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