Question

(vii) At what height the acceleration due to gravity is 25% of
that at the surface of the Earth, in terms of radius of the
Earth.​

Answer 1

Answer:

Explanation:

25/100g=g[1-2h/r]

25/100=[1-2h/r].

25/100=r-2h/r

25r=100r-200h

100r-25r=200h

75r=200h

75r/200=h

3r/8=h

Answer 2

To find :

• Height (h) at which acceleration due to gravity(g) will be 25% of that at the surface of the Earth.

Solution :

• Let the acceleration due to gravity at the surface of the earth be g₀.
• We know that, acceleration due to gravity at the surface of the earth is given by :

        g₀ = $\frac{GM}{R^{2} }$          

      where G is the universal gravitational constant, M is the mass of the Earth and R is the radius of the Earth.

• Acceleration due to gravity at a height h above the earth's surface is given by :

       $g = \frac{GM}{(R+h)^{2} }$

• As per the given condition, g is 25% of g₀.
• ∴ g = 0.25 ×g₀

• ∴ $\frac{GM}{(R+h)^{2} } = \frac{1}{4} (\frac{GM}{R^{2} })$

• ∴ $\frac{R^{2} }{(R+h)^{2} } = \frac{1}{4}$

• ∴ $\frac{R}{R+h} = \frac{1}{2}$
• ∴ $2R = R + h$
• ∴ $R = h$

Answer : The acceleration due to Gravity will be 25% of that at the surface of the Earth at height of R.