Question

When a falling meteoroid is at a distance above the earth's surface of 2.60 times the earth's radius, what is its acceleration due to the earth's gravitation?

Answer 1

Answer: g =0.76 m/s^2

Explanation: The gravitational acceleration is calculated with the equation.

g = G*M/r^

Where G is a constant of gravitation, M is the mass of the Earth and r is the distance of the meteoroid to the earth (to the center of the earth).

G = 6.67408×10^-11 m^3*kg^-1*s^-2

Earth radius = 6371km

M = 5.972 × 10^24 kg

and we know that r = 2.6 times the radius of the earth + radius of the earth, so r = 3.6*6371km

and we need to write this in meters, so r = 3.6*6371*1000m

and g = (6.67408×10^-11 m^3*kg^-1*s^-2)( 5.972 × 10^24 kg)/(3.6*1000*6371m)^2

= 0.76 m/s^2

Answer 2

The gravitational acceleration at any distance r is given by

`g=  (GM)/(r^2)`

where G is the gravitational constant, M the Earth's mass and r is the distance measured from the center of the Earth.

The Earth's radius is
`r_e=6.37 \cdot 10^6 m`, so the meteoroid is located at a distance of:

`r=r_e+2.60 r_e =3.60 r_e =  2.29 \cdot 10^7 m`

And by substituting this value into the previous formula, we can find the value of g at that altitude:

`g=  (GM)/(r^2) =  ((6.67 \cdot 10^(-11) m^3 kg^(-1) s^(-2))(5.97 \cdot 10^(24) kg))/((2.29 \cdot 10^7 m)^2) =0.75 m/s^2`