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?

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skyluke89 skyluke89 · Answer 1 · 2018-04-17T16:08:33+02:00

The gravitational acceleration at any distance r is given by

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

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 [tex]r_e=6.37 \cdot 10^6 m[/tex], so the meteoroid is located at a distance of:

[tex]r=r_e+2.60 r_e =3.60 r_e = 2.29 \cdot 10^7 m[/tex]

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

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

facundo3141592 facundo3141592 · Answer 2 · 2019-11-30T13:05:17+01:00

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