Physics: Circular Motion?

What is the acceleration due to gravity at an altitude of 1.00 × 106 m above the earth’s surface? Note: the radius of the earth is 6.38 × 106 m.

Please help. Thanks.

1 Answer

Relevance
  • 2 months ago
    Favorite Answer

    Answer: a = 7.32 m/s²

    Basic formula

    F = m a

    this is 2nd Law of motion

    a = F/m

    where F is the force of gravity

    F = G M m / r²

    a = F/m = G M / r² ......... (1)

    G is gravitational constant

    M is the mass of the Earth

    r is the distance from the center of the Earth

    We are not given G and M, but remember that the acceleration due to gravity on the surface of the Earth is g=9.8 m/s². As in the previous case

    g = G M / R² ............ (2)

    where distance from the center of the Earth is actually the radius of the earth R

    Divide (1) by (2) to rid of GM

    a / g = R² / r²

    a = g R² / r²

    r = R + h

    where h is the altitude above the Earth's surface. Substitute this in the previous equation

    a = g R² / (R+h)²

    a = g (R/(R+h))²

    a = 9.8 * (6.38*10⁶/(6.38*10⁶+1*10⁶))²

    10⁶ cancels out

    = 9.8 * (6.38 / 7.38)²

    = 9.8 * 0.864²

    = 7.32 m/s²

     

Still have questions? Get your answers by asking now.