A particle of mass m is projected with a velo | vedclass

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

Question

(D) By the law of conservation of energy, the total energy at the surface equals the total energy at the maximum height $r$ where velocity is zero.$-\

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) By the law of conservation of energy, the total energy at the surface equals the total energy at the maximum height $r$ where velocity is zero.$-\frac{GMm}{R} + \frac{1}{2} m v^2 = -\frac{GMm}{r}$Given $v = k V_{e}$ and $V_{e} = \sqrt{\frac{2GM}{R}}$, we have $v^2 = k^2 \frac{2GM}{R}$.Substituting this into the energy equation:$-\frac{GMm}{R} + \frac{1}{2} m \left( k^2 \frac{2GM}{R} \right) = -\frac{GMm}{r}$Dividing by $GMm$:$-\frac{1}{R} + \frac{k^2}{R} = -\frac{1}{r}$$\frac{1}{r} = \frac{1}{R} - \frac{k^2}{R} = \frac{1-k^2}{R}$$r = \frac{R}{1-k^2}$Since $r = R + h$, the maximum height $h$ is:$h = r - R = \frac{R}{1-k^2} - R = R \left( \frac{1 - (1-k^2)}{1-k^2} \right) = \frac{R k^2}{1-k^2}$.

$A$ particle of mass $m$ is projected with a velocity $v = k V_{e}$ $(k < 1)$ from the surface of the earth. $(V_{e} = \text{escape velocity})$. The maximum height above the surface reached by the particle is:

Similar Questions

The ratio of the acceleration due to gravity on two planets $P_1$ and $P_2$ is $K_1$. The ratio of their respective radii is $K_2$. The ratio of their respective escape velocities is

The escape velocity for a planet whose mass is six times the mass of Earth and whose radius is twice the radius of Earth will be (where $V_{e}$ is the escape velocity from the Earth).

If $g$ is the acceleration due to gravity at the earth's surface and $r$ is the radius of the earth,the escape velocity for a body to escape out of the earth's gravitational field is

$A$ body is projected vertically upwards from the surface of the Earth with a velocity equal to one-third of the escape velocity. The maximum height attained by the body will be $...... \ km$. (Take radius of Earth $R = 6400 \ km$ and $g = 10 \ m/s^2$)

What is the escape velocity for the surface of a black hole?

Vedclass Test Series

Exam Paper Generator

Online Exam Module