Physics Question?

A satellite is orbiting the earth at an altitude where the gravitational acceleration is 8.7 m/s^2. What is the orbital speed?

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  • 2 months ago

    Re = 6,371,000 m

    GMe/Re² = 9.8 and GMe/r² = 8.7  and GMe/8.7 = 9.8Re²/8.7 = r²

    Re*√(9.8/8.7) = r = 6,761,780 m  

    mV²/r = m*8.7 and V² = 8.7*r = 8.7*6,761,780 = (7670m/s)²

    V = 7670m/s

  • 2 months ago

    Gravitational acceleration on the surface of the Earth

    g₀ = 9.80 m/s²

    radius of Earth

    r₀ = 6371 km = 6.371x10⁶ m

    Because

    g₀ = GM/r₀²

    g = GM/r²

    dividing the two

    g₀ / g = r² / r₀²

    r = r₀ √(g₀ / g)

    r = 6371 √(9.80 / 8.7) = 6762 km

    When r is known, equalize gravitational and centripetal acceleration

    g = v² / r

    v = √(g r) = √(8.7 * 6.762x10⁶ ) = 7670 m/s

  • Anonymous
    2 months ago

    a = v^2 / r

    then

    v^2 = a r

    also

    a = g (R / r)^2

    then

    r = R (g / a)^1/2

    v^2 = R (a g)^1/2 = 6,37x10^6*(8,7*9,8)^1/2 = 5,88x10^7 m^2/s^2

    v = 7,67x10^3 m/s

  • 2 months ago

    first find height above center of earth

    F = G m₁m₂/r²

    F/m = g = GM/r² = 8.7

    r² = GM/8.7

    r = √(GM/8.7) = 6.76e6 m

       (only 390 km above the surface)

    V = √(GM/R)

    V = √(3.98e14/6.76e6)

    V = 7670 m/s

    Satellite motion, circularV = √(GM/R)T = 2π√[R³/GM]  T is period of satellite in sec  V = velocity in m/s  G = 6.673e-11 Nm²/kg²  M is mass of central body in kg  R is radius of orbit in m

    earth mass M 5.974e24 kg

    earth radius    6,371 km = 6.37e6 meters

    Gravitational attraction in newtons

    F = G m₁m₂/r²

    G = 6.674e-11 m³/kgs²

    m₁ and m₂ are the masses of the two objects in kg

    r is the distance in meters between their centers

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