Question

A meteor is falling towards the earth. If the mass and radius of the earth are 6×10^24 kg and 6.4×10^3 km respectively. Find the hieght of the meteor where its acceleration due to gravity is 4m/s^2.

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Answer 1

h = 4004.349 km

Step-by-step explanation:

Given,

The mass of the Earth, M = 6 x 10²⁷ km

The radius of the Earth, R = 6.4 x 10³ km

At what height the acceleration of the meteor is, h = 4 m/s²

The acceleration due to gravity at the surface of the Earth is given by the formula,

g = GM/r²

At height 'h' acceleration is given by the formula,

`g_(h)` = GM/
`R_(h)^(2)`

Where,

`R_(h)^(2)` = R + h

`R_(h)^(2)` = GM/
`g_(h)`

`R_(h)^(2)` = 6.673 x 10⁻¹¹ X 6 x 10²⁴ / 4

= 1.00095 x 10¹⁴

`R_(h)` = 10004748.87 m

= 10004.749 km

From above,

`R_(h)` = R + h

h =
`R_(h)` - R

= 10004.749 km - 6400 km

= 4004.349 km

Hence, the hieght of the meteor where its acceleration due to gravity is 4 m/s² is, h = 4004.349 km