posted by donna .
n object is dropped from rest from a height
5.8 × 106 m above the surface of the earth.
If there is no air resistance, what is its speed
when it strikes the earth? The acceleration
of gravity is 9.81 m/s2 and the radius of the
Earth is 6.37 × 106 m.
Answer in units of km/s.
That is a large height compared to the radius of the earth (R = 6.37 x 10^6 m), so the strength of gravity changes as it falls. It is 9.81 m/s^2 near the surface of the earth, but less at higher altitudes, following an inverse square law..
If the Earth were much larger than the drop height H, the answer would be
V = sqrt(2gH) = 10,670 m/s = 10.67 km/s
The correct answer is, however
V^2/2 = GMe*[1/R - 1/(R+H)]
= GMe/R^2 *[R - R/(1 + H/R)]
where Me is the mass of the earth and G is the universal constant of gravity.
Make use of the fact that GMe/R^2 = g
V^2 = 2*g*R*[1 - 1/(1 + (H/R))]
= 2*g* (6.37*10^6)[0.464]
V = 7.62 km/s