A body of mass m is thrown vertically upward | vedclass

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

Question

(D) According to the law of conservation of energy, the total energy at the surface of the Earth is equal to the total energy at infinity (or a very l

Vedclass · Accepted Answer

$\sqrt{2} v_e$

Vedclass · Answer

(D) According to the law of conservation of energy, the total energy at the surface of the Earth is equal to the total energy at infinity (or a very large distance where the potential energy is zero).Total energy at the surface: $E_i = K_i + U_i = \frac{1}{2} m(\sqrt{3} v_e)^2 - \frac{GMm}{R}$.We know that the escape velocity $v_e = \sqrt{\frac{2GM}{R}}$, so $v_e^2 = \frac{2GM}{R}$, which implies $\frac{GM}{R} = \frac{v_e^2}{2}$.Substituting this into the energy equation: $E_i = \frac{1}{2} m(3 v_e^2) - m(\frac{v_e^2}{2}) = \frac{3}{2} m v_e^2 - \frac{1}{2} m v_e^2 = m v_e^2$.At a very large distance, the potential energy is $0$, so the final energy is $E_f = \frac{1}{2} m v^2$.Equating $E_i = E_f$: $m v_e^2 = \frac{1}{2} m v^2$.$v^2 = 2 v_e^2 \implies v = \sqrt{2} v_e$.

$A$ body of mass $m$ is thrown vertically upward with speed $\sqrt{3} v_e$, where $v_e$ is the escape velocity of a body from the Earth's surface. The final velocity of the body is

Similar Questions

If $M$ is the mass of a planet and $R$ is its radius,then in order to become a black hole (where $c$ is the speed of light),which condition must be satisfied?

The mass and radius of the earth and moon are $M, R$ and $m, r$ respectively. The distance between their centers is $d$. The minimum velocity with which a particle of mass $m_0$ should be projected from the midpoint between them so that it could reach infinity is

The escape velocity for a planet whose radius is $1.7 \times 10^6 \ m$ and acceleration due to gravity is $1.7 \ m s^{-2}$ is

The escape velocity of a body from the Earth is about $11.2 \, km/s$. Assuming the mass and radius of the Earth to be about $81$ and $4$ times the mass and radius of the Moon,respectively,the escape velocity in $km/s$ from the surface of the Moon will be .......

The escape velocity for the earth is $v_e$. The escape velocity for a planet whose radius is four times and density is nine times that of the earth,is (in $,v_e$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module