Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 months ago

Physics help please?

The Hubble Space Telescope orbits the Earth every 95 minutes, taking breathtaking images of our universe. Calculate its altitude above the Earth’s 

Surface???

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  • NCS
    Lv 7
    3 months ago
    Favorite Answer

    For "orbit", centripetal acceleration = gravitational acceleration, or

    ω²r = (2π/T)²r = 4π²r / T² = v²/r = GM/r²

    where the first four terms are all expressions for centripetal acceleration

    and G = Newton's gravitational constant = 6.674e−11 N·m²/kg²

    and M = mass of central body

    and r = orbit radius

    and T = period

    and v = orbit velocity

    4π²r / T² = GM/r²

    r³ = GMT² / 4π²

    r³ = 6.674e−11N·m²/kg² * 5.98e24kg * (95*60 s)² / 4π²

    r³ = 3.28e28 m³

    r = 6.90e6 m

    altitude h = r - Re = 6.90e6m - 6.371e6m = 5.3e5 m

    or 530 km

    If your reference tables have a different value for the mass and/or radius of the Earth, use it.

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