A body of mass m is projected with velocity l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to the law of conservation of energy,the total energy at the surface of the earth is equal to the total energy at the maximum height $h$

Vedclass · Accepted Answer

$\frac{R}{1-\lambda^{2}}$

Vedclass · Answer

(B) According to the law of conservation of energy,the total energy at the surface of the earth is equal to the total energy at the maximum height $h$ from the center of the earth.Total energy at surface = Total energy at height $h$$-\frac{GMm}{R} + \frac{1}{2}m(\lambda v_{e})^{2} = -\frac{GMm}{h} + 0$Since the escape velocity $v_{e} = \sqrt{\frac{2GM}{R}}$,we have $v_{e}^{2} = \frac{2GM}{R}$.Substituting this into the equation:$-\frac{GMm}{R} + \frac{1}{2}m\lambda^{2}(\frac{2GM}{R}) = -\frac{GMm}{h}$$-\frac{GMm}{R} + \frac{\lambda^{2}GMm}{R} = -\frac{GMm}{h}$Dividing both sides by $-GMm$:$\frac{1}{R} - \frac{\lambda^{2}}{R} = \frac{1}{h}$$\frac{1-\lambda^{2}}{R} = \frac{1}{h}$$h = \frac{R}{1-\lambda^{2}}$

Similar Questions

$A$ planet has twice the radius but the mean density is $\frac{1}{4}^{th}$ as compared to Earth. What is the ratio of the escape velocity from Earth to that from the planet?

If the acceleration due to gravity $g$ doubles and the radius of earth becomes half that of the present value,then the value of escape velocity is (Assume,$g=10 \ m/s^2$ and radius of earth,$R=6400 \ km$)

$A$ body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth,the maximum height attained by the body from the surface of the earth is

The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $V_{e}$,then the escape velocity from the planet is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module