Question

When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration due to the Earth's gravitation

Answer 1

Answer:

g = 0.85 m$s^{-2}$

Explanation:

g = $\frac{GM}{h^{2} }$

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x $10^{-11}$ N$m^{2}$$kg^{-2}$), M is the mass of the earth (5.972 x $10^{24}$ kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

  = 3.40 x 6371 km

h = 21661.4 km

  = 21661400 m

Thus,

g = $\frac{6.674*10^{-11}*5.972*10^{24} }{(21661400)^{2} }$

  = $\frac{3.9857 *10^{14} }{4.6922*10^{14} }$

  = 0.84944

g = 0.85 m$s^{-2}$

The acceleration due to the Earth's gravitation is 0.85 m$s^{-2}$.