A bullet is fired vertically upwards with velocity v from the surface of a spherical planet- When it reaches its maximum height- its acceleration due to the planet-s gravity is 1-4 th of its value at the surface of the planet-If the escape velocity from the planet is- v sec - v - N - then the value of N is - Ignore energy loss due to atmosphere

At the maximum height, g' = 1/4 g ⇒GM/r^2 = 1/4GM/R^2 ⇒ r = 2R By conservation of mechanical energy, GMm/R + 1/2mv^2 = GMm/2R× 0 ⇒1/2 mv^2 = GMm/2R ⇒ ...

You visited us 72 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Distance

A bullet is f...

Question

A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet’s gravity is $\frac{1}{4}$ th of its value at the surface of the planet.If the escape velocity from the planet is, $v_{s e c} = v \sqrt{N},$ then the value of N is . (Ignore energy loss due to atmosphere)

Open in App

Solution

At the maximum height,

$g^{'} = \frac{1}{4} g \Rightarrow \frac{G M}{r^{2}} = \frac{1}{4} \frac{G M}{R^{2}}$

$\Rightarrow r = 2 R$

By conservation of mechanical energy,

$- \frac{G M m}{R} + \frac{1}{2} m v^{2} = - \frac{G M m}{2 R} \times 0$

$\Rightarrow \frac{1}{2} m v^{2} = \frac{G M m}{2 R}$

$\Rightarrow v = \sqrt{\frac{G M}{R}}$

$v_{e s c} = \sqrt{\frac{2 G M}{R}} = v \sqrt{2} = v \sqrt{N}$

$\Rightarrow N = 2.$

Suggest Corrections

Similar questions

Q. A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is

V

. Due to the rotation of planet about its axis the acceleration due to gravity

g

at equator is

\frac{1}{2}

g

at poles. The escape velocity of a particle on the pole of planet in terms of

V

Q. A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is

V

. Due to the rotation of a planet about its axis the acceleration due to gravity

g

at equator is

\frac{1}{2}

g

at poles. The escape velocity of a particle on the pole of a planet in terms of

V

Q. A rocket starts vertically upward with speed

v_{0}

. Show that its speed v at height h is given by

v_{0}^{2} - v^{2} = \frac{2 h g}{1 + \frac{h}{R}}

where R is the radius of the earth and g is acceleration due to gravity at earth's surface. Deduce an expression for maximum height reached by a rocket fired with speed

0.9

times the escape velocity.

Q. In a hypothetical uniform and spherical planet of mass

M

and radius

R

, a tunnel is dug radially from its surface to its centre as shown. The minimum energy required to carry a unit mass from its centre to the surface is

k m g R

. Find value of

k

. Acceleration due to gravity at the surface of the planet is

g

Q. The radius of a planet is

1 / 4^{t h}

R_{e}

and its acc, due to gravity is

2 g

. What would be the value of escape velocity on the planet, if escape velocity on earth is

v_{e}

Join BYJU'S Learning Program

At the maximum height, g′=14g⇒GMr2=14GMR2 ⇒r=2R By conservation of mechanical energy, −GMmR+12mv2=−GMm2R×0 ⇒12mv2=GMm2R ⇒v=√GMR vesc=√2GMR=v√2=v√N ⇒N=2.