Question

( 2.) At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the earth is 6.4x10^6 metres and the acceleration due to gravity 9.8m/s^2 on the surface of the earth

Answer 1

\[G M/r^2=4.9\]

Answer 2

G=6.67300 × 10-11

Answer 3

(6.67300 × 10-11)(M)/(R+6.4*10^6)=4.9

Answer 4

where m = mass of the earth

Answer 5

You don't need those things. Since it's an inverse square law, the acceleration due to gravity will be half as great when the radial distance from the earth increases by a factor of sqrt(2). Therefore,
\[R' = \sqrt{2} R \approx 9.05\times10^6 m\]

Answer 6

And then you must subtract off the old radius, because it wants height above the earths surface, so that's
\[9.05\times10^6m - 6.4 \times 10^6m = 2.65 \times 10^6 m \]