A rocket of mass M is launched vertically fro | vedclass

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

Question

(C) By the law of conservation of energy,the total energy at the surface of the earth equals the total energy at the maximum height $h$.At the surface

Vedclass · Accepted Answer

$R / \left( \frac{2gR}{V^2} - 1 \right)$

Vedclass · Answer

(C) By the law of conservation of energy,the total energy at the surface of the earth equals the total energy at the maximum height $h$.At the surface: $E_i = K_i + U_i = \frac{1}{2}MV^2 - \frac{GM_eM}{R}$At maximum height $h$: $E_f = K_f + U_f = 0 - \frac{GM_eM}{R+h}$Equating $E_i = E_f$: $\frac{1}{2}MV^2 - \frac{GM_eM}{R} = - \frac{GM_eM}{R+h}$$\frac{1}{2}V^2 = GM_e \left( \frac{1}{R} - \frac{1}{R+h} \right)$Using $g = \frac{GM_e}{R^2}$,we have $GM_e = gR^2$.$\frac{V^2}{2} = gR^2 \left( \frac{R+h-R}{R(R+h)} \right) = \frac{gR^2h}{R(R+h)} = \frac{gRh}{R+h}$$\frac{V^2}{2gR} = \frac{h}{R+h}$$V^2(R+h) = 2gRh$$V^2R + V^2h = 2gRh$$V^2R = h(2gR - V^2)$$h = \frac{V^2R}{2gR - V^2}$Dividing numerator and denominator by $V^2$: $h = \frac{R}{\frac{2gR}{V^2} - 1}$

$A$ rocket of mass $M$ is launched vertically from the surface of the earth with an initial speed $V$. Assuming the radius of the earth to be $R$ and negligible air resistance,the maximum height attained by the rocket above the surface of the earth is

Similar Questions

What is the change in the gravitational potential energy of an object when it moves away from the Earth and when it comes closer to the Earth?

$A$ body of mass $m$ is placed on the earth's surface. It is taken from the earth's surface to a height $h = 3R$. The change in gravitational potential energy of the body is

$A$ body of mass $m$ falls from a height $R$ above the surface of the earth,where $R$ is the radius of the earth. What is the velocity attained by the body on reaching the ground? (Acceleration due to gravity on the surface of the earth is $g$)

An object is projected vertically upward from the earth's surface with a velocity $u = \sqrt{Rg}$,where $R$ is the radius of the earth and $g$ is the acceleration due to gravity on the earth's surface. Find the maximum height reached by the object.

Vedclass Test Series

Exam Paper Generator

Online Exam Module