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4 June, 12:08

Modeled after the two satellites problem (slide 12 in Lecture13, Universal gravity), imagine now if the period of the satellite is exact 88.59 hours, the earth mass is 5.98 x 1024 kg, and the radius of the earth is 3958.8 miles, what is the distance of the satellite from the surface of the earth in MILES? Use G=6.67x 10 - 11 Nm2/kg2. Your answer could be a large number on the order of tens of thousands, just put in the raw number you get, for example, 12345.67.

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  1. 4 June, 12:34
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    R = 6.3456 10⁴ mile

    Explanation:

    For this exercise we will use Newton's second law where force is gravitational force

    F = m a

    The satellite is in a circular orbit therefore the acceleration is centripetal

    a = v² / r

    Where the distance is taken from the center of the Earth

    G m M / r² = m v² / r

    G M / r = v²

    The speed module is constant, let's use the uniform motion relationships, with the length of the circle is

    d = 2π r

    v = d / t

    The time for a full turn is called period (T)

    Let's replace

    G M / r = (2π r / T) ²

    r³ = G M T²²2 / 4π²

    r = ∛ (G M T² / 4π²)

    We have the magnitudes in several types of units

    T = 88.59 h (3600 s / 1h) = 3.189 10⁵ s

    Re = 6.37 10⁶ m

    Let's calculate

    r = ∛ (6.67 10⁻¹¹ 5.98 10²⁴ (3,189 10⁵) ²/4π²)

    r = ∛ (1.027487 10²⁴)

    r = 1.0847 10⁸ m

    This is the distance from the center of the Earth, the distance you want the surface is

    R = r - Re

    R = 108.47 10⁶ - 6.37 10⁶

    R = 102.1 10⁶ m

    Let's reduce to miles

    R = 102.1 10⁶ m (1 mile / 1609 m)

    R = 6.3456 10⁴ mile
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