Question

can anyone please help me​

Answer 1

Gravitational field at a point is the force of attraction acting on a unit mass placed at that point.

So the gravitational field at a distance $\sf{\dfrac{R_e}{4}}$ from center is given by,

$\longrightarrow\sf{F=\dfrac{GM'}{\left(\dfrac{R_e}{4}\right)^2}}$

Or,

$\longrightarrow\sf{F=\dfrac{16GM'}{\left(R_e\right)^2}\quad\quad\dots(1)}$

where $\sf{M'}$ is the mass of the earth when its radius decreases to $\sf{\dfrac{R_e}{4}}$ but not the actual mass of the earth of radius $\sf{R_e.}$

But the density of earth won't change even the radius is reduced, since earth is uniform. Hence we have,

$\longrightarrow\sf{\rho=\dfrac{M'}{\dfrac{4}{3}\pi\left(\dfrac{R_e}{4}\right)^3}}$

$\longrightarrow\sf{\rho=\dfrac{48M'}{\pi\left(R_e\right)^3}}$

$\longrightarrow\sf{M'=\dfrac{\pi\rho(R_e)^3}{48}}$

since earth is spherical in shape.

Hence (1) becomes,

$\longrightarrow\sf{F=\dfrac{16G\cdot\pi\rho(R_e)^3}{48\left(R_e\right)^2}}$

$\longrightarrow\sf{\underline{\underline{F=\dfrac{\pi G\rho\,R_e}{3}}}}$

Hence (1) is the answer.

Answer 2

first of all use formula of density to calculate value of mass

after calculating mass put it in the formula

g=GMr/R³

where M is mass, r is Height from surface of earth and R is radius of earth

to calculate volume use the formula

V= 4πR³/3