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Satellite motion - determining planets' masses

  1. Feb 19, 2014 #1
    1. Determine the ratio of the masses of the planets Earth and Mars by using only information about their orbital periods and orbital radii about the sun. Assume the planets can be treated as points with mass and assume circular orbits.



    2. Relevant equations
    Gravitational force: [tex] F_g = G m_1 m_2 / R^2 [/tex]
    where [itex]G[/itex] is the gravitational constant and [itex]R[/itex] is the distance between the two masses.

    Tangential velocity in uniform circular motion: [tex] v = 2\piR/T [/tex]
    where T is the period.


    Centripetal acceleration in uniform circular motion: [tex] a_c = v^2/R = 4(\pi^2)R/T^2 [/tex]

    3. The attempt at a solution
    I know centripetal force is provided by the force of gravity, so:
    [itex] F_g = G m_1 m_2 / R^2 [/itex] [itex] = m_1a_c [/itex]

    I can express centripetal acceleration in terms of radius and period and shown above, but since [itex] m_1 [/itex] appears on both sides of the equation! I conclude that the mass of an orbiting object has nothing to do with its orbital period or it's orbital radius.
    This professor is notorious for accidentally giving problems that can't actually be done, and I suspect that this is one. Am I missing something?

    In case that wasn't clear, my problem is that the mass of an orbiting body can't be expressed in terms of its orbital radius and period. Thus I can't determine the ratio of the masses of two orbiting bodies using only this information.
     
    Last edited: Feb 19, 2014
  2. jcsd
  3. Feb 19, 2014 #2

    Andrew Mason

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    You are right. The mass of the central gravitating body (sun) can be measured using this technique but not the mass of the orbiting body.

    AM
     
  4. Feb 19, 2014 #3

    D H

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    That is an incorrect conclusion. Do the math right and you will find that the orbital period is ##2\pi\sqrt{\frac {a^3}{G(M_{\text{sun}}+M_{\text{planet}})}}## .

    However, since the Sun is about 333,000 times as massive as is the Earth, you are going to need extremely precise data to solve this problem (i.e., you'll need the semimajor axes and periods of Earth's orbit and of Mar's orbit to at least eight places). That extremely precise data doesn't exist as far as I can tell.

    There is one missing piece of data. What you can infer from the periods and radii is the ratio ##\frac{M_{\text{sun}}+M_{\text{mars}}}{M_{\text{sun}}+M_{\text{earth}}}##. You'll need extremely precise data to properly determine that ratio. To be able to compute the ratio of the mass of the Earth to that of Mars, you'll also need the ratio of the mass of Earth (or of Mars) to that of the Sun.
     
  5. Feb 19, 2014 #4

    Andrew Mason

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    I think it is implicit in the question that the reduced mass is not an issue. In order to know the distance of the sun from the sun-earth and sun-mars centres of mass we would have to know the mass of each planet, which is what the question is asking us to find.

    AM
     
  6. Feb 19, 2014 #5

    D H

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    I beg to differ. The only way to come anywhere close to answering the question as posed is to account for how the mass of a planet changes the period of that planet's orbit.
     
  7. Feb 19, 2014 #6

    Andrew Mason

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    But before you can do that you would first have to know how the mass of the planet compares to the mass of the sun. It is implicit in the question that the mass of the sun is so much larger that this factor can be ignored (i.e. Kepler's third law applies). The question also says to assume a circular orbit for each planet, which implies that the centre of rotation is the same for each planet ie. which can only be the centre of mass of the sun.

    This question is similar to asking someone to calculate the mass of an earth satellite from its orbital radius and period. We have known since Galileo that all bodies at the same distance from the earth centre accelerate at the same rate regardless of mass. So we know that the mass of the satellite does not affect the orbital period at a given radius (or the radius for a given orbital period).

    AM
     
    Last edited: Feb 19, 2014
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