If the density of the earth is doubled keeping the radius constant,find the new acceleration due to gravity (in $m/s^2$)? $(g = 9.8 \ m/s^2)$

If a particle takes $t$ seconds less and acquires a velocity of $v \text{ m/s}$ more in falling through the same distance on two planets,where the accelerations due to gravity are $2g$ and $8g$ respectively,then:

The acceleration due to gravity becomes $\left(\frac{g}{2}\right)$ ($g =$ acceleration due to gravity on the surface of the earth) at a height equal to

At what height over the earth's pole,the free fall acceleration decreases by one percent? (Assume the radius of earth to be $6400 \, km$)

Choose the correct alternative:
$(a)$ Acceleration due to gravity increases/decreases with increasing altitude.
$(b)$ Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to be a sphere of uniform density).
$(c)$ Acceleration due to gravity is independent of mass of the Earth/mass of the body.
$(d)$ The formula $-G M m(1 / r_{2}-1 / r_{1})$ is more/less accurate than the formula $m g(r_{2}-r_{1})$ for the difference of potential energy between two points $r_{2}$ and $r_{1}$ distance away from the centre of the Earth.

If the value of $g$ at the surface of the earth is $9.8 \, m/s^2$,then the value of $g$ at a place $480 \, km$ above the surface of the earth will be .......... $m/s^2$. (Radius of the earth is $6400 \, km$)