Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 decade ago

physics question? how do you find apogee and perigee velocity?

A satellite is placed into a circular low-earth orbit at an altitude of h km. A perigee rocket motor is fired to increase its speed by 2.4 km/s. This places the satellite into an elliptical orbit. At the apogee of this elliptical orbit, another motor (the apogee motor) is fired to increase the speed of the satellite to place it into a circular orbit at that altitude. What is the speed change required from the apogee motor in km/s?

h[km] = 599;

Can you please show working. Cheers, Ian.

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  • 1 decade ago
    Favorite Answer

    The total energy of the satellite is constant and is just the sum of the potential and kinetic energies.

    Kinetic = (1/2)m*v^2

    Potential = -GMm/r

    Total Energy = (1/2)m*v^2 - GMm/r

    This total energy is also equal to 1/2 the potential energy at a distance equal to the semi-major axis of the orbit.

    Energy = -(1/2)GMm/a

    Equating these two gives:

    (1/2)m*v^2 - GMm/r = -(1/2)GMm/a

    Now solve this for "a" since we do not have the apogee distance.

    v^2 = GM(2/r - 1/a)

    v2/(GM) = 2/r - 1/a

    1/a = 2/r - v2/(GM)

    a = 1/[2/r - v^2/(GM)]

    For the elliptical orbit the changes in velocity occur at perigee and apogee. Call these distances from the center of the earth r1 and r2.

    2*a = r1 + r2

    a = (r1 + r2)/2

    (r1 + r2)/2 = 1/[2/r - v^2/(GM)]

    r2 = 2/[2/r - v^2/(GM)] - r1

    Get r2. We do this at the apogee point where r=r1, which we know, and we also know v.

    G = 6.67300x10^(-11) m^3 kg^-1 s^-2

    M = 5.9742x10^24 kg

    h = 5.99x10^5 m

    r1 = Re + h = 6378 + 599 km = 6.977x10^6 m

    v = (circular orbit speed) + 2.4 km/s

    v = SQRT(GM/r) = 9960 m/s

    Using all this gives:

    r2 = 4.58x10^7 m

    We now know the altitude for the new circular orbit and can calculate the circular orbital velocity using SQRT(GM/r). We get:

    V(circular new) = 2948 m/s

    To get the speed change at this point we need to know the speed in the eliiptical orbit at apogee. We can use the equation for speed as a function of r from above and remembering that r = r2 at this point.

    v^2 = GM(2/r2 - 1/a)

    v = 1515 m/s

    The speed change is then 2948 - 1515 = 1433 m/s = 1.43 km/s

    So the speed change required to put the satellite into a circular orbit is 1.43 km/sec.

    Check. The formulas for change in velocity are:

    Dv1 = SQRT(GM/r1)[SQRT(2*r2/(r1+ r2)) - 1]

    Dv2 = SQRT(GM/r2)[1 - SQRT(2*r2/(r1+ r2))]

    If you use what was calculated and the given data these work out to:

    Dv1 = 2.4 and Dv2 = 1.43

    So the answer looks good.

  • Wendy
    Lv 4
    5 years ago

    According to your stated rule, it takes 2 fps for every mile reduction in the altitude. Ergo, dV = 2.0*dA = 2.0*60 = 120 fps In reality, the perigee and apogee will not change by the same amount unless they are equal to start with.

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