Home
Class 12
PHYSICS
Maximum height reached by a rocket fired...

Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from earth's surface is:

A

`R//2`

B

`16R//9`

C

`R//3`

D

`R//8`

Text Solution

Verified by Experts

The correct Answer is:
C
`v=50/100 V_(e)=1/2 sqrt((2GM)/R)`
Applying energy conservation
`implies-(GMm)/R+1/2mv^(2)=-(GMm)/((R+h))`
`v^(2)=(2GM)/R-(2GM)/(R+h)implies1/4.(2GM)/R=2GM(1/R-1/(R+h))`
`implies 1/(4R)=h/(R(R+h))`
`implies R+h=4himpliesh=R//3`
Promotional Banner

Similar Questions

Explore conceptually related problems

maximum height reached by a bullet fired vertically upward with a speed equal to 50% if the escape velocity from earth's surface is (R is radius of earth):

What is the maximum height attained by a body projected with a velocity equal to one- third of the escape velocity from the surface of the earth? (Radius of the earth=R)

An artificial satellite moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth's surface, will be at a height_______________ km.

Write the formula for the escape velocity of a body from the earth's surface.

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the escape velocity from the earth of radius R. The height of the satellite above the surface of the earth is

The escape velocity of a body from the surface of earth is

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)=(2hg)/(1+h/R) where R is the radius of the earth and g is acceleration due to gravity at earth's suface. Deduce an expression for maximum height reachhed by a rocket fired with speed 0.9 times the escape velocity.