Question

6) Height at which value of g
becomes 1/4 th to that on earth (R =
radius of earth) *
OR
O
2R
O (3/2)R
4R​

Answer 1

Answer:

Let the height from the surface of earth be 'h' and 'r' be the radius of the earth.

According to Newton's Law of Gravitation,

$\text{Gravitational Force} = \dfrac{GMm}{r^2}$

Hence Acceleration due to gravity is nothing but Force divided by Mass. Hence we get:

$\implies g = \dfrac{F_g}{m}\\\\\\\implies \boxed{\bf{g = \dfrac{GM}{r^2}}} \hspace{10} ...(1)$

Now the gravity at a height (r + h), we get the value of 'g' to be g/4.

$\implies \dfrac{g}{4} = \dfrac{GM}{(r+h)^2} \hspace{10} ...(2)$

Dividing (1) by (2) we get:

$\implies \dfrac{g}{g/4} = \dfrac{GM}{r^2} \times \dfrac{(r+h)^2}{GM}\\\\\\\implies \dfrac{4}{1} = \dfrac{(r+h)^2}{r^2}\\\\\\\implies \dfrac{(2)^2}{(1)^2} = \dfrac{(r+h)^2}{(r^2)}$

Taking Square root on both sides we get:

$\implies \dfrac{2}{1} = \dfrac{r+h}{r}\\\\\\\text{Cross multiplying we get:}\\\\\implies 2r = r + h\\\\\implies 2r - r = h\\\\\implies \boxed{ \bf{ h = r}}$

Hence the height from the centre of earth where gravity is (g/4) is given as:

⇒ r + h

⇒ r + r = 2r

Hence the required answer is 2r.