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Homework Help: Spacecraft altitude and orbit problem

  1. Apr 10, 2006 #1
    A spacecraft (with its cargo, a satellite) is circling the Earth at a
    constant speed and an altitude of 600km. The combined mass of the craft and
    the satellite is 1.20×10^5 kg (kilograms).

    At phi = 0, the captain ejects the 1.10×10^4 kg satellite out the
    back. The speed of the satellite relative to the spacecraft at the time of
    separation is 100 m/s (meters per second).

    When the spacecraft reaches phi = pi, what will its altitude above the
    Earth be?
  2. jcsd
  3. Apr 10, 2006 #2
    Here is my attempt at answering the above question:

    Applying the Principle of Conservation of Linear Momentum:

    0 = -(1.10×10^4)×100 + (1.20×10^5)×V

    Hence V = 9.167 m/s, where V is the relative speed of the spacecraft after
    ejecting the satellite.

    I assume that 'phi = 0' means that the spacecraft is currently at the
    perigee of an elliptic orbit and I need to find its altitude when phi = pi
    (at the apogee).

    I believe I am meant to use some or all of the following equations:

    r = l/(1 + e cos(phi))

    l = L^2/GMm^2

    e = AL^2/GMm^2

    1/r = Acos(phi) + GMm/L^2


    G = 6.67×10^-11 Nm^2kg^-2

    M = 5.98×10^24 kg

    radius of Earth: 6.37×10^6 m

    However I have no idea what to do next. Please advise me. THANK YOU.
  4. Apr 10, 2006 #3
    Well, you have an equation that says what the radial position is if you know the angular momentum. Conservation of momentum gives you the new orbital speed of the spacecraft.

    So how do you find the new angular momentum of the spacecraft?

  5. Apr 10, 2006 #4


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    Homework Helper

    You have a couple problems.

    One, you calculated the difference in velocity between the initial speed of the spacecraft and final speed, but still don't know the speed of either.

    Secondly, your second equation can be simplified by using the 'specific angular momentum per unit mass' (the mass is squared in both the numerator and denominator, canceling out). The easier equation is:

    [tex]r = \frac{\frac {h^2}{\mu}}{1 + e cos \phi}[/tex]

    [tex]\mu[/tex] is just the universal gravitational constant times the mass of the Earth (no sense multiplying two constants by each other over and over - in fact, the WGS-84 value for it is 3.986004418 x 10^14 m^3/sec^2).
    Edit: Since you're using rounded numbers for G and M, your number won't match the WGS-84 value. In fact, with a radius of 6370 km, you're probably using a different datum than WGS-84. The WGS-84 radius is 6378.137 km.

    I mention this only because the specific angular momentum is also equal to the cross product of the radius vector and the velocity vector - or, algebraically, the radius times the speed times the angle between the two vectors if the origins coinided. Since you started out in a circular orbit, the velocity is always perpendicular to the radius. That means you can use the specific angular momentum to determine your initial speed. Or, if you left the mass in to get your real angular momentum, you could still find your linear momentum, and then divide by your mass (it's just easier to leave the mass out).

    Once you have your initial speed, you could use conservation of linear momentum to determine the final speed (actually, since you already calculated how much the spacecraft sped up, you could just add that to your initial speed).

    [tex](1.20*10^5)(v_i)=(1.1*10^4)(v_f-100) + (1.09*10^5)(v_f)[/tex]

    From there, you should get a better start.
    Last edited: Apr 10, 2006
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