(D) The orbital velocity of the satellite at height $h$ is given by $v = \sqrt{\frac{GM}{R+h}}$. Since $h \ll R$,we can approximate $R+h \approx R$. T

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(D) The orbital velocity of the satellite at height $h$ is given by $v = \sqrt{\frac{GM}{R+h}}$. Since $h \ll R$,we can approximate $R+h \approx R$. Thus,$v = \sqrt{\frac{GM}{R}}$.Using the relation $g = \frac{GM}{R^2}$,we get $GM = gR^2$. Substituting this,$v = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}$.The escape velocity $v'$ required to escape the Earth's gravitational field from height $h$ is $v' = \sqrt{\frac{2GM}{R+h}}$.Again,using $h \ll R$,$v' = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.The minimum increase in speed required is $\Delta v = v' - v$.$\Delta v = \sqrt{2gR} - \sqrt{gR} = \sqrt{gR}(\sqrt{2} - 1)$.

Class 11 Physics Gravitation Mix Examples-Gravitation

Medium

$A$ satellite is revolving in a circular orbit at a height $h$ from the Earth's surface,such that $h \ll R$,where $R$ is the radius of the Earth. Assuming that the effect of the Earth's atmosphere can be neglected,the minimum increase in the speed required so that the satellite could escape from the gravitational field of the Earth is:

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A
$\sqrt{2gR}$
B
$\sqrt{gR}$
C
$\sqrt{\frac{gR}{2}}$
D
$\sqrt{gR}(\sqrt{2}-1)$

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Two planets $A$ and $B$ having masses $m_1$ and $m_2$ move around the sun in circular orbits of $r_1$ and $r_2$ radii respectively. If angular momentum of $A$ is $L$ and that of $B$ is $3L$,the ratio of time period $\left(\frac{T_A}{T_B}\right)$ is:

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Let $\omega$ be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth $d$ below the earth's surface at a pole $(d << R)$. The value of $d$ is

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Two point masses of mass $4m$ and $m$ respectively,separated by a distance $d$,are revolving under their mutual force of attraction. The ratio of their kinetic energies is:

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Match List-$I$ with List-$II$:
List-$I$ List-$II$
$(A)$ Kinetic energy of planet $(1)$ $-\frac{GMm}{a}$
$(B)$ Gravitational potential energy of Sun-planet system $(2)$ $\frac{GMm}{2a}$
$(C)$ Total mechanical energy of planet $(3)$ $\frac{GM}{r}$
$(D)$ Escape energy at the surface of planet for unit mass object $(4)$ $-\frac{GMm}{2a}$

(Where $a=$ radius of planet orbit,$r=$ radius of planet,$M=$ mass of Sun,$m=$ mass of planet)
Choose the correct answer from the options given below:

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The orbital velocity of a body near the surface of a planet $A$ is equal to the escape velocity of a body from the planet $B$. If the masses of planets $A$ and $B$ are the same,the ratio of their radii is

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List-$I$	List-$II$
$(A)$ Kinetic energy of planet	$(1)$ $-\frac{GMm}{a}$
$(B)$ Gravitational potential energy of Sun-planet system	$(2)$ $\frac{GMm}{2a}$
$(C)$ Total mechanical energy of planet	$(3)$ $\frac{GM}{r}$
$(D)$ Escape energy at the surface of planet for unit mass object	$(4)$ $-\frac{GMm}{2a}$

Similar Questions

Two planets $A$ and $B$ having masses $m_1$ and $m_2$ move around the sun in circular orbits of $r_1$ and $r_2$ radii respectively. If angular momentum of $A$ is $L$ and that of $B$ is $3L$,the ratio of time period $\left(\frac{T_A}{T_B}\right)$ is:

Two point masses of mass $4m$ and $m$ respectively,separated by a distance $d$,are revolving under their mutual force of attraction. The ratio of their kinetic energies is:

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