Question

At what Height above the surface of the earth will the Acceleration due to gravity be 25% of it's value on the surface of the earth? Assume that the radius of the earth is 6400km.


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Answer 1

Let the value of acceleration due to gravity be g.

Also,

The mass of the earth = M

Radius given, R = 6400 km

We know,

g = GM / R² ...(1)

Now let the height above the earth's surface at which the value of g is 25% or 1/4 of the value on the surface of the earth be x m

Now,

g = GM / r²

⇒ g' = GM / (r + x)² ...(2)

Here, The ratio of g' and g is 1/4 because g' is 25% of g

⇒ g' / g = { GM / (r + x)² } / { GM / r² }

⇒ 1/4 = r² / (r + x)²

Raising the power 1/2 both sides,

⇒ 1/2 = r / (r + x)

⇒ r + x = 2r

⇒ x = 2r - r

⇒ x = r

⇒ x = 6400km

Hence, At height of 6400 km from the earth's surface, The value of g will be 25% of itself.

Some Information :-

◉ Gravitational force is a force of attraction between two objects which have mass M1 and M2 respectively located at a distance d between them.

Then the magnitude of Force will be:

F = G × M1 × M2 / d²

• S.I Unit = Newton
• G = Gravitational constant, 6.67 × 10^-11 Nm²/kg²

◉ Acceleration due to gravity of a body is the rate of attraction at which a body would be attracted towards it.

• It is denoted by g

g = GM/r²

Where,

• G = Gravitational constant
• M = Mass of the body
• r = Radius of the body

Answer 2

At a height equal to radius of earth

g' = g/4

so 6400+6400 = 12800 Km

hope it helps

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