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    Gravitation: Force, motion of satellite, Kepler's law

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    1. A 8754-kg satellite is orbiting planet Y at an altitude of 1.69x10^6 meters above its surface and with an orbital period of 15.2 hours. If the planet has a mass of 8.19x10^24 kg, then determine the radius of planet Y.

    2. Suppose the earth had another moon which was 1.69 times as far from the center of the earth as our own moon. Determine the orbital period of this moon if our own moon has a period of 27.32 days(2 pt.)

    3. A 1692-kg object is located between the earth and the moon. The mass of the earth is 6.0x10^24 kg and the mass of the moon is 7.40x10^22 kg. The distance from the earth's center to the moon's center is 3.84x10^8meters . (4 pts.)The distance from the object to the moon's center is 2 times longer than the distance from the object to the earth's center. What is the net force on the object between the earth and moon.

    © BrainMass Inc. brainmass.com October 4, 2022, 4:30 pm ad1c9bdddf
    https://brainmass.com/physics/planets/gravitation-force-motion-satellite-keplers-law-60766

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    A 8754-kg satellite is orbiting planet Y at an altitude of 1.69x10^6 meters above its surface and with an orbital period of 15.2 hours. If the planet has a mass of 8.19x10^24 kg, then determine the radius of planet Y.

    Let the mass of the planet M, that of satellite n, height of the satellite h and the radius of the satellite is R.
    The distance of the satellite from the center of the planet will be R + h
    Newton's law as gives force of gravity on the satellite
    F = GMm/(R + h)2
    This force will behave as the centripetal force, required to move the satellite on the circular track of radius R + h hence
    GMm/(R + h)2 = m2(R + h) = m(2/T)2(R + h)
    Where is the angular velocity of the satellite, which is 2/T. T is time period.
    The above relation gives us

    gives R + h = 34.61*106 m
    hence the radius of the planet
    R = 34.61*106 - 1.69*106 = 32.92*106 m

    --------------------------------------------
    Suppose the earth had another moon which was 1.69 times as far from the center of the earth as our own moon. Determine the orbital period of this moon if our own moon has a period of 27.32 days(2 pt.)

    According to Kepler's law

    T2  r3
    Where T is the time period for the planet(satellite) and r is the radius of orbit of planet(satellite)
    Or
    Or
    Gives T2 = 60.02 days.

    --------------------------
    A 1692-kg object is located between the earth and the moon. The mass of the earth is 6.0x10^24 kg and the mass of the moon is 7.40x10^22 kg. The distance from the earth's center to the moon's center is 3.84x10^8meters . (4 pts.)The distance from the object to the moon's center is 2 times longer than the distance from the object to the earth's center. What is the net force on the object between the earth and moo

    Let the distance of the object from center of earth is r and that from moon is 2r.
    The total distance is
    r + 2r = 3r = 3.84*108 m
    gives r = 1.28*108 m

    The resultant gravitation field at the object is given by

    where is the unit vector in the positive x direction.

    substituting the values we get

    m/s2.

    Hence the magnitude of the force on the object is

    F = mg = 1692*0.0243 =41.19 N and it is towards the earth.
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    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 4, 2022, 4:30 pm ad1c9bdddf>
    https://brainmass.com/physics/planets/gravitation-force-motion-satellite-keplers-law-60766

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