$A$ geostationary satellite is revolving around the $Earth$. To make it escape from the gravitational field of the $Earth$,its velocity must be increased by ........ $\%$

Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface of the earth is ($R$ - Radius of the earth).

$A$ particle of mass $m_0$ is projected from the midpoint of the line joining two fixed particles,each of mass $m$. If the separation between the fixed particles is $l$,the minimum velocity of projection of the particle so as to escape is equal to:

An object $A$ of mass '$m$' is located at a point '$P$' at distances '$r$' and '$2r$' from two planets $B$ and $C$ of masses '$M$' and '$6M$' respectively,as shown in the figure. If the escape speed of the object $A$ from point '$P$' due to the gravitational influence of only planet $B$ is $5 \ km/s$,then the escape speed of the object $A$ from point '$P$' due to the gravitational influence of both the planets is . . . . . . $km/s$.

The ratio of accelerations due to gravity $g_{1}:g_{2}$ on the surfaces of two planets is $5:2$ and the ratio of their respective average densities $\rho_{1}:\rho_{2}$ is $2:1$. What is the ratio of respective escape velocities $v_{1}:v_{2}$ from the surface of the planets?

The initial velocity $v_{i}$ required to project a body vertically upward from the surface of the earth to reach a height of $10 R$,where $R$ is the radius of the earth,may be described in terms of escape velocity $v_{e}$ such that $v_{i} = \sqrt{\frac{x}{y}} \times v_{e}$. The value of $x$ will be ...... .