111.

Mass of a planet is 10 times that of earth and radius is 2 times that of earth. If at the earth, the escape velocity of a satellite is V_es(e), then escape. velocity at the planet is :

• $\sqrt{5}$v_es(e)

• 5v_es(e)

• 2$\sqrt{5}$ v_es(e)

• 10 v_es(e)

A rocket is accelerated with speed v= 2$\sqrt{g{R}_e}$, near earth surface and then it moves upward. At far distance from the earth surface, the speed of the rocket will
be:

• $\sqrt{g{R}_e/2}$

• $\sqrt{g{R}_e}$

• (2-$\sqrt{2}$)$\sqrt{g{R}_e}$

• $\sqrt{2g{R}_e}$

A satellite is in a circular orbit round the earth at an altitude R above the earth's surface,where R is the radius of the earth. If g is the acceleration due to gravity on the surface of the earth the speed of the satellite is

• $\sqrt{2\mathrm{Rg}}$

• $\sqrt{\mathrm{Rg}}$

• $\sqrt{\mathrm{Rg}/2}$

• $\sqrt{\mathrm{Rg}/4}$

Two satellites S₁ and S₂ revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 h and 8 h respectively. The radius of the orbit of S₁ is 10⁴ km. The speed of S₂ relative to S₁ when they are closest, in km/h is

• 10⁴ $\mathrm{\pi }$

• 2 × 10⁴ $\mathrm{\pi }$

• 10⁴ $\frac{\mathrm{\pi }}{2}$

• 4 × 10⁴ $\mathrm{\pi }$

If the height of a satellite from the earth is negligible in comparison to the radius of the earth R, the orbital velocity of the satellite is

• gR

• $\frac{\mathrm{gR}}{2}$

• $\sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$

• $\sqrt{\mathrm{gR}}$

A satellite is revolving round the earth in an elliptical orbit. Its speed

• will be same at all points of the orbit

• will be maximum when it is at maximum distance from earth

• will be maximum when its distance from the earth will be minimum

• goes on increasing or decreasing continuous depending upon the mass of the satellite

A satellite of mass m, revolving round a planet of mass M in a circular orbit of radius r, then orbital velocity of the satellite is

• $\frac{\mathrm{mg}}{\mathrm{r}}$

• $\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$

• $\sqrt{\frac{2\mathrm{GM}}{\mathrm{r}}}$

• $\frac{\mathrm{GM}}{\mathrm{r}}$

If the radius of the earth is reduced by 1% without changing its mass. The change in g will be

• 2% decrease

• 2% increases

• 1% decrease

• 1% increases

If the change in the value of g at a height h above the surface of the earth is the same as at a depth x below in surface, then

• x = h²

• x = 0.5 h

• x = 2h

• x = h

Radius of orbit of satellite of earth is R its kinetic energy is proportional to

• $\frac{1}{\mathrm{R}}$

• $\frac{1}{\sqrt{\mathrm{R}}}$

• R

• $\frac{1}{{\mathrm{R}}^3}$