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

If a planet of mass $6.4 \times 10^{23} \ kg$ can be compressed into a sphere such that the escape velocity from its surface is $8 \times 10^4 \ m/s$,then what should be the radius of the sphere (in $km$)? (Gravitational constant,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$)

If the escape velocity on Earth is $11.2 \text{ km/s}$, what is its value for a planet having double the radius and $8$ times the mass of Earth (in $\text{ km/s}$)?

Given the mass of the moon is $1/81$ of the mass of the earth and its radius is $1/4$ of the earth's radius. If the escape velocity on the earth's surface is $11.2 \, km/s$,the value of the escape velocity on the surface of the moon is ......... $km/s$.

$A$ body is projected up with a velocity equal to $\frac{3}{4}$ of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth $= R$)

The escape velocity for a body projected vertically upwards from the surface of earth is $11.2 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ......... $km/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module