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

Respuesta :

Answer:

g = 0.85 m[tex]s^{-2}[/tex]

Explanation:

g = [tex]\frac{GM}{h^{2} }[/tex]

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x [tex]10^{-11}[/tex] N[tex]m^{2}[/tex][tex]kg^{-2}[/tex]), M is the mass of the earth (5.972 x [tex]10^{24}[/tex] 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 = [tex]\frac{6.674*10^{-11}*5.972*10^{24} }{(21661400)^{2} }[/tex]

  = [tex]\frac{3.9857 *10^{14} }{4.6922*10^{14} }[/tex]

  = 0.84944

g = 0.85 m[tex]s^{-2}[/tex]

The acceleration due to the Earth's gravitation is 0.85 m[tex]s^{-2}[/tex].