A spacecraft is in a circular Earth orbit at an altitude of 6000 km . By how much will its altitude decrease if it moves to a new circular o

Question

A spacecraft is in a circular Earth orbit at an altitude of 6000 km . By how much will its altitude decrease if it moves to a new circular orbit where (a) its orbital speed is 10% higher or (b) its orbital period is 10% shorter?

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Ngọc Hoa 1 year 2021-07-26T14:33:52+00:00 1 Answers 62 views 0

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    2021-07-26T14:35:35+00:00

    Answer:

    a) 2148 km = 2150 km

    b) 840 km

    Explanation:

    The force keeping the satellite in circular motion is the force given by Netwon’s gravitational law

    Centripetal force = (mv²/r)

    Force due to Newton’s law of gravitation = (GMm/r²)

    where m = mass of satellite

    M = mass of the earth

    G = Gravitational constant

    v = velocity of the satellite

    r = radius of circular orbit

    (mv²/r) = (GMm/r²)

    v² = (GM/r)

    Meaning that the square of the velocity of orbit is inversely proportional to the radius of circular orbit. (Since G and M are constants)

    v² = (k/r)

    when v = v₀, r = 6000 + 6378 = 12,378 km (the radius of orbit = 6000 km + radius of the earth)

    v₀² = (k/12,378)

    K = 12378v₀²

    When the velocity increases by 10%, v₁ = 1.1v₀, the square of the new velocity = (1.1v₀)² = 1.21v₀² and the new radius of orbit = r₁

    1.21v₀² = (k/r₁)

    r₁ = (k/1.21v₀²)

    Recall, k = 12378v₀²

    r₁ = 12378v₀² ÷ 1.21v₀² = 10,229.75 km

    10,229.75 km = (10,229.75 – 6378) km altitude above the Earth’s surface

    New altitude of orbit = 3851.75 km

    Decrease in altitude = 6000 – 3851.75 = 2148 km

    b) The period of orbit is related to the radius of orbit through Kepler’s Law

    T² ∝ R³

    T² = kR³

    When the period of orbit is T₀, Radius of orbit = R₀ = (6000 + 6378) = 12378 km (Earth’s radius = 6378 km)

    T₀² = kR₀³

    T₀² = k(12378)³

    k = (T₀²) ÷ (12378)³

    When the period reduces by 10%, T₁ = 0.90T₀ and the new radius of orbit = R₁

    T₁² = kR₁³

    (0.90T₀)² = kR₁³

    0.81T₀² = kR₁³

    R₁³ = (0.81T₀²) ÷ k

    Recall, k = (T₀²)/(12378)³

    R₁³ = (0.81T₀²) ÷ [(T₀²)/(12378)³]

    R₁³ = 1,536,160,005,663.1

    R₁ = ∛(1,536,160,005,663.1) = 11,538.4 km

    New Altitude = R₁ – (Radius of the Earth)

    = 11,538.4 – 6378 = 5160.4 km

    Decrease in altitude = 6000 – 5160.4 = 839.6 km = 840 km

    Hope this Helps!!!

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