A space station is at a height equal to the r | vedclass

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

Question

(C) The escape velocity $v$ at a distance $r$ from the center of the Earth is given by the formula $v = \sqrt{\frac{2GM}{r}}$.For the surface of the E

Vedclass · Accepted Answer

$1/\sqrt{2}$

Vedclass · Answer

(C) The escape velocity $v$ at a distance $r$ from the center of the Earth is given by the formula $v = \sqrt{\frac{2GM}{r}}$.For the surface of the Earth,the distance $r = R$,so the escape velocity is $V_{E} = \sqrt{\frac{2GM}{R}}$.For a space station at a height $h = R$ above the surface,the distance from the center of the Earth is $r = R + h = R + R = 2R$.The escape velocity $v_{s}$ at the space station is $v_{s} = \sqrt{\frac{2GM}{2R}} = \sqrt{\frac{GM}{R}}$.Comparing $v_{s}$ with $V_{E}$:$v_{s} = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}} = \frac{1}{\sqrt{2}} V_{E}$.Thus,the escape velocity on the space station is $\frac{1}{\sqrt{2}}$ times $V_{E}$.

$A$ space station is at a height equal to the radius of the Earth. If $V_{E}$ is the escape velocity on the surface of the Earth,the escape velocity on the space station is __ times $V_{E}$.

Similar Questions

The escape velocity from the surface of the Earth is $V_e$. The escape velocity from the surface of a planet whose mass and radius are $3$ times those of the Earth will be:

$A$ very small groove is made in the earth,and a particle of mass $m_0$ is placed at $\frac{R}{2}$ distance from the centre. Find the escape speed of the particle from that place.

How much energy is necessary for a body of $500 \, kg$ to escape from the Earth? $[g = 9.8 \, m/s^2$,radius of Earth $R = 6.4 \times 10^6 \, m]$

$A$ satellite with kinetic energy $E_k$ is revolving around the Earth in a circular orbit. How much more kinetic energy should be given to it so that it may just escape into outer space?

Vedclass Test Series

Exam Paper Generator

Online Exam Module