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Math Help - gravity question

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    gravity question

    A satellite is called stationary or synchronous if its period (i.e. the time required for
    one orbital revolution) is one day. What is the height above the Earth’s surface at which a synchronous satellite needs to be placed?
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  2. #2
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    Quote Originally Posted by poirot View Post
    A satellite is called stationary or synchronous if its period (i.e. the time required for
    one orbital revolution) is one day. What is the height above the Earth’s surface at which a synchronous satellite needs to be placed?
    Kepler's 3rd law of planetary motion (derived from Newton's Law of Universal Gravitation = centripetal force) ...

    T^2 = \dfrac{4\pi^2}{GM_e} \cdot R^3<br />

    R = R_e + h

    you need to solve for the orbital radius, R , then determine the value of h
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    A similar reasoning: The centripetal acceleration v^2/(R+h)=\omega^2(R+h) is provided by the gravity, so it equals GM/(R+h)^2. Here R is the Earth's radius, M is the Earth's mass, h is the satellite's height above the Earth’s surface, v is the satellite's velocity, \omega is the satellite's angular velocity and G is the gravitational constant. The angular velocity is given.
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