Question

At what altitude the value of 'g' would become one fourth than on the surface of the Earth?

Answer 1

Answer:

It is R

Explanation:

R is the radius of Earth

Answer 2

Answer:

The altitude of the earth will be equal to the radius of the earth i.e.6.3781×10⁶m.

Explanation:

At a particular height, the acceleration due to gravity is given as,

$g'=\frac{GM}{(r+h)^2}$             (1)

And  g at the surface of the earth is given as,

$g=\frac{GM}{r^2}$              (2)

Where,

g'=acceleration due to gravity at a particular height

M=mass of the earth

h=height above the surface of the earth

G=universal gravitational constant

r=radius of the earth=6.3781×10⁶m

Now, according to the question,

$g'=\frac{g}{4}$               (3)

By using equations (1) and (2) in (3) we get;

$\frac{GM}{(r+h)^2}=\frac{1}{4}\times\frac{GM}{r^2}$

$\frac{r^2}{(r+h)^2}=\frac{1}{4}$

$\frac{r}{r+h}=\frac{1}{2}$

$2r=r+h$

$r=h=6.3781\times 10^6m$

Hence, the altitude of the earth will be equal to the radius of the earth i.e.6.3781×10⁶m.