Question

Calculate the effective value of g, the acceleration of gravity, at (a) 6400 m, and (b) 6400 km, above the earth's surface.

Answer 1

a) 9.8m/s² as height is not so effective.
b) Gm/(r+6400)² is the formula
G=6.67*10-¹¹, m=mass of earth, r=radius of
earth.

Answer 2

Answer:

9.79395 m/s²

2.44228 m/s²

Explanation:

m = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 km

a) at h = 6400 m

$g=G\frac{M}{r^2}\\\Rightarrow g=6.67\times 10^{-11}\frac{5.972\times 10^{24}}{(6371\times 10^3+6400)^2}\\\Rightarrow g=9.79395\ m/s^2$

g = 9.79395 m/s²

b) at h = 6400000 m

$g=G\frac{M}{r^2}\\\Rightarrow g=6.67\times 10^{-11}\frac{5.972\times 10^{24}}{(6371\times 10^3+6400000)^2}\\\Rightarrow g=2.44228\ m/s^2$

g = 2.44228 m/s²