QBManager

PHXI08:GRAVITATION

359727 A body of mass $m$ is moving in a circular orbit of radius $R$ about a planet of mass $M$ . At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $\frac{R}{2}$ and the other mass, in a circular orbit of radius $\frac{3 R}{2}$ . The difference bewteen the final and initial total energies is:

1

+ \frac{G m}{6 R}

2

- \frac{G M m}{2 R}

3

- \frac{G M m}{6 R}

4

\frac{G M m}{2 R}

Explanation:

$E_{1} = - \frac{G M m}{2 R}$
$\begin{aligned} E_{f} & = - \frac{G M (m / 2)}{2 (\frac{R}{2})} - \frac{G M (m / 2)}{2 (\frac{3 R}{2})} = - \frac{2 G M m}{3 R} \\ = - \frac{4 G M m}{6 R} = - \frac{2 M m}{3 R} \\ E_{f} - E_{i} & = \frac{G M m}{R} (- \frac{2}{3} + \frac{1}{2}) = - \frac{G M m}{6 R} \end{aligned}$

JEE - 2018

PHXI08:GRAVITATION

359728 A satellite of $10^{3} k g$ mass is revolving in circular orbit of radius $2 R$ . If $\frac{10^{4} R}{6} J$ energy is supplied to the satellite, it would revolve in a new circular orbit of radius(use $g = 10 m / s^{2},$ $R =$ radius of earth)

1

6 R

2

2.5 R

3

3 R

4

4 R

Explanation:

Mass of satellite, $m = 10^{3} k g, r = 2 R$
Total energy of satellite is given by
$E_{T} = \frac{- G M m}{R}$
$∴$ Total energy $= \frac{- G M m}{2 (2 R)}$
Now energy is given $= \frac{10^{4} R}{6}$
So, $\frac{- G M m}{4 R} + \frac{10^{4} R}{6} = \frac{- G M m}{2 r}$
( $r$ is new radius, $G M = g R^{2}$ )
$\Rightarrow \frac{- m g R}{4} + \frac{10^{4} R}{6} = - \frac{m g R^{2}}{2 r}$
$\Rightarrow \frac{- 10^{3} \times 10 R}{4} + \frac{10^{4} R}{6} = \frac{- 10^{3} \times 10 \times R^{2}}{2 r}$
$\Rightarrow r = 6 R$

JEE - 2024

PHXI08:GRAVITATION

359729 A satellite is orbiting with areal velocity $v_{a}$ in circular orbit. At what height from the surface of the earth, it is rotating, if the radius of earth is $R$ ?

1

\frac{4 v_{a}^{2}}{g R^{2}} - R

2

\frac{2 v_{a}^{2}}{g R^{2}} - R

3

\frac{v_{a}^{2}}{g R^{2}} - R

4

\frac{v_{a}^{2}}{2 g R^{2}} - R

Explanation:

Areal velocity $= \frac{L}{2 m}$
$v_{a} = \frac{1}{2 m} m v r$
$\begin{aligned} \Rightarrow v_{a} = \frac{v r}{2} = \frac{r}{2} \sqrt{\frac{G M}{r}} = \frac{r}{2} \sqrt{\frac{g R^{2}}{r}} = \frac{1}{2} \sqrt{g R^{2} r} \\ \Rightarrow \frac{4 v_{a}^{2}}{g R^{2}} = r = R + h \Rightarrow h = \frac{4 v_{a}^{2}}{g R^{2}} - R . \end{aligned}$

PHXI08:GRAVITATION

359730 The figure shows the variation of energy with the orbit radius of a body in circular planetary motion. Find the correct statement about the curves $A, B$ and $C$
supporting img

1

A

shows the kinetic energy,

B

the potential energy and

C

the total energy of the system

2

C

shows the total energy,

B

the kinetic energy and

A

the potential energy system

3

C

and

A

are kinetic and potential energies respectively and

B

is the total energy of the system

4

A

and

B

are the kinetic and potential energies respectively and

C

is the total energy of the system

Explanation:

Let $m$ is the mass of the body and $M$ is the mass of the earth. The orbital velocity is
$\begin{aligned} v_{o} = \sqrt{\frac{G M}{r}} \\ K . E = \frac{1}{2} m v_{o}^{2} = \frac{G M m}{2 r} \\ P . E = \frac{- G M m}{r} \\ T . E = \frac{- G M m}{2 r} \end{aligned}$

PHXI08:GRAVITATION

359731 Two satellites $P$ and $Q$ of same mass are revolving near the earth surface in the equitorial plane. The satellite $P$ moves in the direction of rotation of earth whereas $Q$ moves in the opposite direction. The ratio of their kinetic energies with respect to a frame attached to earth will be -

1

{(\frac{7437}{8363})}^{2}

2

{(\frac{8363}{7437})}^{2}

3

(\frac{7437}{8363})

4

(\frac{8363}{7437})

Explanation:

Linear velocity of earth,
$v_{e} = \frac{2 π R_{e}}{T_{e}} = \frac{6.28 \times 6.4 \times 10^{6}}{24 \times 3600} = 463 m / s$
Orbital velocity,
According to question, $v_{0} = \sqrt{R_{e} g} = 7.9 \times 10^{3} m / s$
According to question,
$v_{K P} = v_{0} - v_{e} = 7900 - 463 = 7437 m / s$
$v_{K Q} = v_{0} + v_{e} = 7900 + 463 = 8363 m / s$
$∴ \frac{E_{K Q}}{E_{K P}} = \frac{v_{K Q}^{2}}{v_{K P}^{2}} = {(\frac{8363}{7437})}^{2}$

PHXI08:GRAVITATION

359727 A body of mass $m$ is moving in a circular orbit of radius $R$ about a planet of mass $M$ . At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $\frac{R}{2}$ and the other mass, in a circular orbit of radius $\frac{3 R}{2}$ . The difference bewteen the final and initial total energies is:

1

+ \frac{G m}{6 R}

2

- \frac{G M m}{2 R}

3

- \frac{G M m}{6 R}

4

\frac{G M m}{2 R}

Explanation:

$E_{1} = - \frac{G M m}{2 R}$
$\begin{aligned} E_{f} & = - \frac{G M (m / 2)}{2 (\frac{R}{2})} - \frac{G M (m / 2)}{2 (\frac{3 R}{2})} = - \frac{2 G M m}{3 R} \\ = - \frac{4 G M m}{6 R} = - \frac{2 M m}{3 R} \\ E_{f} - E_{i} & = \frac{G M m}{R} (- \frac{2}{3} + \frac{1}{2}) = - \frac{G M m}{6 R} \end{aligned}$

JEE - 2018

PHXI08:GRAVITATION

359728 A satellite of $10^{3} k g$ mass is revolving in circular orbit of radius $2 R$ . If $\frac{10^{4} R}{6} J$ energy is supplied to the satellite, it would revolve in a new circular orbit of radius(use $g = 10 m / s^{2},$ $R =$ radius of earth)

1

6 R

2

2.5 R

3

3 R

4

4 R

Explanation:

Mass of satellite, $m = 10^{3} k g, r = 2 R$
Total energy of satellite is given by
$E_{T} = \frac{- G M m}{R}$
$∴$ Total energy $= \frac{- G M m}{2 (2 R)}$
Now energy is given $= \frac{10^{4} R}{6}$
So, $\frac{- G M m}{4 R} + \frac{10^{4} R}{6} = \frac{- G M m}{2 r}$
( $r$ is new radius, $G M = g R^{2}$ )
$\Rightarrow \frac{- m g R}{4} + \frac{10^{4} R}{6} = - \frac{m g R^{2}}{2 r}$
$\Rightarrow \frac{- 10^{3} \times 10 R}{4} + \frac{10^{4} R}{6} = \frac{- 10^{3} \times 10 \times R^{2}}{2 r}$
$\Rightarrow r = 6 R$

JEE - 2024

PHXI08:GRAVITATION

359729 A satellite is orbiting with areal velocity $v_{a}$ in circular orbit. At what height from the surface of the earth, it is rotating, if the radius of earth is $R$ ?

1

\frac{4 v_{a}^{2}}{g R^{2}} - R

2

\frac{2 v_{a}^{2}}{g R^{2}} - R

3

\frac{v_{a}^{2}}{g R^{2}} - R

4

\frac{v_{a}^{2}}{2 g R^{2}} - R

Explanation:

Areal velocity $= \frac{L}{2 m}$
$v_{a} = \frac{1}{2 m} m v r$
$\begin{aligned} \Rightarrow v_{a} = \frac{v r}{2} = \frac{r}{2} \sqrt{\frac{G M}{r}} = \frac{r}{2} \sqrt{\frac{g R^{2}}{r}} = \frac{1}{2} \sqrt{g R^{2} r} \\ \Rightarrow \frac{4 v_{a}^{2}}{g R^{2}} = r = R + h \Rightarrow h = \frac{4 v_{a}^{2}}{g R^{2}} - R . \end{aligned}$

PHXI08:GRAVITATION

359730 The figure shows the variation of energy with the orbit radius of a body in circular planetary motion. Find the correct statement about the curves $A, B$ and $C$
supporting img

1

A

shows the kinetic energy,

B

the potential energy and

C

the total energy of the system

2

C

shows the total energy,

B

the kinetic energy and

A

the potential energy system

3

C

and

A

are kinetic and potential energies respectively and

B

is the total energy of the system

4

A

and

B

are the kinetic and potential energies respectively and

C

is the total energy of the system

Explanation:

Let $m$ is the mass of the body and $M$ is the mass of the earth. The orbital velocity is
$\begin{aligned} v_{o} = \sqrt{\frac{G M}{r}} \\ K . E = \frac{1}{2} m v_{o}^{2} = \frac{G M m}{2 r} \\ P . E = \frac{- G M m}{r} \\ T . E = \frac{- G M m}{2 r} \end{aligned}$

PHXI08:GRAVITATION

359731 Two satellites $P$ and $Q$ of same mass are revolving near the earth surface in the equitorial plane. The satellite $P$ moves in the direction of rotation of earth whereas $Q$ moves in the opposite direction. The ratio of their kinetic energies with respect to a frame attached to earth will be -

1

{(\frac{7437}{8363})}^{2}

2

{(\frac{8363}{7437})}^{2}

3

(\frac{7437}{8363})

4

(\frac{8363}{7437})

Explanation:

Linear velocity of earth,
$v_{e} = \frac{2 π R_{e}}{T_{e}} = \frac{6.28 \times 6.4 \times 10^{6}}{24 \times 3600} = 463 m / s$
Orbital velocity,
According to question, $v_{0} = \sqrt{R_{e} g} = 7.9 \times 10^{3} m / s$
According to question,
$v_{K P} = v_{0} - v_{e} = 7900 - 463 = 7437 m / s$
$v_{K Q} = v_{0} + v_{e} = 7900 + 463 = 8363 m / s$
$∴ \frac{E_{K Q}}{E_{K P}} = \frac{v_{K Q}^{2}}{v_{K P}^{2}} = {(\frac{8363}{7437})}^{2}$

PHXI08:GRAVITATION

359727 A body of mass $m$ is moving in a circular orbit of radius $R$ about a planet of mass $M$ . At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $\frac{R}{2}$ and the other mass, in a circular orbit of radius $\frac{3 R}{2}$ . The difference bewteen the final and initial total energies is:

1

+ \frac{G m}{6 R}

2

- \frac{G M m}{2 R}

3

- \frac{G M m}{6 R}

4

\frac{G M m}{2 R}

Explanation:

$E_{1} = - \frac{G M m}{2 R}$
$\begin{aligned} E_{f} & = - \frac{G M (m / 2)}{2 (\frac{R}{2})} - \frac{G M (m / 2)}{2 (\frac{3 R}{2})} = - \frac{2 G M m}{3 R} \\ = - \frac{4 G M m}{6 R} = - \frac{2 M m}{3 R} \\ E_{f} - E_{i} & = \frac{G M m}{R} (- \frac{2}{3} + \frac{1}{2}) = - \frac{G M m}{6 R} \end{aligned}$

JEE - 2018

PHXI08:GRAVITATION

359728 A satellite of $10^{3} k g$ mass is revolving in circular orbit of radius $2 R$ . If $\frac{10^{4} R}{6} J$ energy is supplied to the satellite, it would revolve in a new circular orbit of radius(use $g = 10 m / s^{2},$ $R =$ radius of earth)

1

6 R

2

2.5 R

3

3 R

4

4 R

Explanation:

Mass of satellite, $m = 10^{3} k g, r = 2 R$
Total energy of satellite is given by
$E_{T} = \frac{- G M m}{R}$
$∴$ Total energy $= \frac{- G M m}{2 (2 R)}$
Now energy is given $= \frac{10^{4} R}{6}$
So, $\frac{- G M m}{4 R} + \frac{10^{4} R}{6} = \frac{- G M m}{2 r}$
( $r$ is new radius, $G M = g R^{2}$ )
$\Rightarrow \frac{- m g R}{4} + \frac{10^{4} R}{6} = - \frac{m g R^{2}}{2 r}$
$\Rightarrow \frac{- 10^{3} \times 10 R}{4} + \frac{10^{4} R}{6} = \frac{- 10^{3} \times 10 \times R^{2}}{2 r}$
$\Rightarrow r = 6 R$

JEE - 2024

PHXI08:GRAVITATION

359729 A satellite is orbiting with areal velocity $v_{a}$ in circular orbit. At what height from the surface of the earth, it is rotating, if the radius of earth is $R$ ?

1

\frac{4 v_{a}^{2}}{g R^{2}} - R

2

\frac{2 v_{a}^{2}}{g R^{2}} - R

3

\frac{v_{a}^{2}}{g R^{2}} - R

4

\frac{v_{a}^{2}}{2 g R^{2}} - R

Explanation:

Areal velocity $= \frac{L}{2 m}$
$v_{a} = \frac{1}{2 m} m v r$
$\begin{aligned} \Rightarrow v_{a} = \frac{v r}{2} = \frac{r}{2} \sqrt{\frac{G M}{r}} = \frac{r}{2} \sqrt{\frac{g R^{2}}{r}} = \frac{1}{2} \sqrt{g R^{2} r} \\ \Rightarrow \frac{4 v_{a}^{2}}{g R^{2}} = r = R + h \Rightarrow h = \frac{4 v_{a}^{2}}{g R^{2}} - R . \end{aligned}$

PHXI08:GRAVITATION

359730 The figure shows the variation of energy with the orbit radius of a body in circular planetary motion. Find the correct statement about the curves $A, B$ and $C$
supporting img

1

A

shows the kinetic energy,

B

the potential energy and

C

the total energy of the system

2

C

shows the total energy,

B

the kinetic energy and

A

the potential energy system

3

C

and

A

are kinetic and potential energies respectively and

B

is the total energy of the system

4

A

and

B

are the kinetic and potential energies respectively and

C

is the total energy of the system

Explanation:

Let $m$ is the mass of the body and $M$ is the mass of the earth. The orbital velocity is
$\begin{aligned} v_{o} = \sqrt{\frac{G M}{r}} \\ K . E = \frac{1}{2} m v_{o}^{2} = \frac{G M m}{2 r} \\ P . E = \frac{- G M m}{r} \\ T . E = \frac{- G M m}{2 r} \end{aligned}$

PHXI08:GRAVITATION

359731 Two satellites $P$ and $Q$ of same mass are revolving near the earth surface in the equitorial plane. The satellite $P$ moves in the direction of rotation of earth whereas $Q$ moves in the opposite direction. The ratio of their kinetic energies with respect to a frame attached to earth will be -

1

{(\frac{7437}{8363})}^{2}

2

{(\frac{8363}{7437})}^{2}

3

(\frac{7437}{8363})

4

(\frac{8363}{7437})

Explanation:

Linear velocity of earth,
$v_{e} = \frac{2 π R_{e}}{T_{e}} = \frac{6.28 \times 6.4 \times 10^{6}}{24 \times 3600} = 463 m / s$
Orbital velocity,
According to question, $v_{0} = \sqrt{R_{e} g} = 7.9 \times 10^{3} m / s$
According to question,
$v_{K P} = v_{0} - v_{e} = 7900 - 463 = 7437 m / s$
$v_{K Q} = v_{0} + v_{e} = 7900 + 463 = 8363 m / s$
$∴ \frac{E_{K Q}}{E_{K P}} = \frac{v_{K Q}^{2}}{v_{K P}^{2}} = {(\frac{8363}{7437})}^{2}$

PHXI08:GRAVITATION

359727 A body of mass $m$ is moving in a circular orbit of radius $R$ about a planet of mass $M$ . At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $\frac{R}{2}$ and the other mass, in a circular orbit of radius $\frac{3 R}{2}$ . The difference bewteen the final and initial total energies is:

1

+ \frac{G m}{6 R}

2

- \frac{G M m}{2 R}

3

- \frac{G M m}{6 R}

4

\frac{G M m}{2 R}

Explanation:

$E_{1} = - \frac{G M m}{2 R}$
$\begin{aligned} E_{f} & = - \frac{G M (m / 2)}{2 (\frac{R}{2})} - \frac{G M (m / 2)}{2 (\frac{3 R}{2})} = - \frac{2 G M m}{3 R} \\ = - \frac{4 G M m}{6 R} = - \frac{2 M m}{3 R} \\ E_{f} - E_{i} & = \frac{G M m}{R} (- \frac{2}{3} + \frac{1}{2}) = - \frac{G M m}{6 R} \end{aligned}$

JEE - 2018

PHXI08:GRAVITATION

359728 A satellite of $10^{3} k g$ mass is revolving in circular orbit of radius $2 R$ . If $\frac{10^{4} R}{6} J$ energy is supplied to the satellite, it would revolve in a new circular orbit of radius(use $g = 10 m / s^{2},$ $R =$ radius of earth)

1

6 R

2

2.5 R

3

3 R

4

4 R

Explanation:

Mass of satellite, $m = 10^{3} k g, r = 2 R$
Total energy of satellite is given by
$E_{T} = \frac{- G M m}{R}$
$∴$ Total energy $= \frac{- G M m}{2 (2 R)}$
Now energy is given $= \frac{10^{4} R}{6}$
So, $\frac{- G M m}{4 R} + \frac{10^{4} R}{6} = \frac{- G M m}{2 r}$
( $r$ is new radius, $G M = g R^{2}$ )
$\Rightarrow \frac{- m g R}{4} + \frac{10^{4} R}{6} = - \frac{m g R^{2}}{2 r}$
$\Rightarrow \frac{- 10^{3} \times 10 R}{4} + \frac{10^{4} R}{6} = \frac{- 10^{3} \times 10 \times R^{2}}{2 r}$
$\Rightarrow r = 6 R$

JEE - 2024

PHXI08:GRAVITATION

359729 A satellite is orbiting with areal velocity $v_{a}$ in circular orbit. At what height from the surface of the earth, it is rotating, if the radius of earth is $R$ ?

1

\frac{4 v_{a}^{2}}{g R^{2}} - R

2

\frac{2 v_{a}^{2}}{g R^{2}} - R

3

\frac{v_{a}^{2}}{g R^{2}} - R

4

\frac{v_{a}^{2}}{2 g R^{2}} - R

Explanation:

Areal velocity $= \frac{L}{2 m}$
$v_{a} = \frac{1}{2 m} m v r$
$\begin{aligned} \Rightarrow v_{a} = \frac{v r}{2} = \frac{r}{2} \sqrt{\frac{G M}{r}} = \frac{r}{2} \sqrt{\frac{g R^{2}}{r}} = \frac{1}{2} \sqrt{g R^{2} r} \\ \Rightarrow \frac{4 v_{a}^{2}}{g R^{2}} = r = R + h \Rightarrow h = \frac{4 v_{a}^{2}}{g R^{2}} - R . \end{aligned}$

PHXI08:GRAVITATION

359730 The figure shows the variation of energy with the orbit radius of a body in circular planetary motion. Find the correct statement about the curves $A, B$ and $C$
supporting img

1

A

shows the kinetic energy,

B

the potential energy and

C

the total energy of the system

2

C

shows the total energy,

B

the kinetic energy and

A

the potential energy system

3

C

and

A

are kinetic and potential energies respectively and

B

is the total energy of the system

4

A

and

B

are the kinetic and potential energies respectively and

C

is the total energy of the system

Explanation:

Let $m$ is the mass of the body and $M$ is the mass of the earth. The orbital velocity is
$\begin{aligned} v_{o} = \sqrt{\frac{G M}{r}} \\ K . E = \frac{1}{2} m v_{o}^{2} = \frac{G M m}{2 r} \\ P . E = \frac{- G M m}{r} \\ T . E = \frac{- G M m}{2 r} \end{aligned}$

PHXI08:GRAVITATION

359731 Two satellites $P$ and $Q$ of same mass are revolving near the earth surface in the equitorial plane. The satellite $P$ moves in the direction of rotation of earth whereas $Q$ moves in the opposite direction. The ratio of their kinetic energies with respect to a frame attached to earth will be -

1

{(\frac{7437}{8363})}^{2}

2

{(\frac{8363}{7437})}^{2}

3

(\frac{7437}{8363})

4

(\frac{8363}{7437})

Explanation:

Linear velocity of earth,
$v_{e} = \frac{2 π R_{e}}{T_{e}} = \frac{6.28 \times 6.4 \times 10^{6}}{24 \times 3600} = 463 m / s$
Orbital velocity,
According to question, $v_{0} = \sqrt{R_{e} g} = 7.9 \times 10^{3} m / s$
According to question,
$v_{K P} = v_{0} - v_{e} = 7900 - 463 = 7437 m / s$
$v_{K Q} = v_{0} + v_{e} = 7900 + 463 = 8363 m / s$
$∴ \frac{E_{K Q}}{E_{K P}} = \frac{v_{K Q}^{2}}{v_{K P}^{2}} = {(\frac{8363}{7437})}^{2}$

PHXI08:GRAVITATION

359727 A body of mass $m$ is moving in a circular orbit of radius $R$ about a planet of mass $M$ . At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $\frac{R}{2}$ and the other mass, in a circular orbit of radius $\frac{3 R}{2}$ . The difference bewteen the final and initial total energies is:

1

+ \frac{G m}{6 R}

2

- \frac{G M m}{2 R}

3

- \frac{G M m}{6 R}

4

\frac{G M m}{2 R}

Explanation:

$E_{1} = - \frac{G M m}{2 R}$
$\begin{aligned} E_{f} & = - \frac{G M (m / 2)}{2 (\frac{R}{2})} - \frac{G M (m / 2)}{2 (\frac{3 R}{2})} = - \frac{2 G M m}{3 R} \\ = - \frac{4 G M m}{6 R} = - \frac{2 M m}{3 R} \\ E_{f} - E_{i} & = \frac{G M m}{R} (- \frac{2}{3} + \frac{1}{2}) = - \frac{G M m}{6 R} \end{aligned}$

JEE - 2018

PHXI08:GRAVITATION

359728 A satellite of $10^{3} k g$ mass is revolving in circular orbit of radius $2 R$ . If $\frac{10^{4} R}{6} J$ energy is supplied to the satellite, it would revolve in a new circular orbit of radius(use $g = 10 m / s^{2},$ $R =$ radius of earth)

1

6 R

2

2.5 R

3

3 R

4

4 R

Explanation:

Mass of satellite, $m = 10^{3} k g, r = 2 R$
Total energy of satellite is given by
$E_{T} = \frac{- G M m}{R}$
$∴$ Total energy $= \frac{- G M m}{2 (2 R)}$
Now energy is given $= \frac{10^{4} R}{6}$
So, $\frac{- G M m}{4 R} + \frac{10^{4} R}{6} = \frac{- G M m}{2 r}$
( $r$ is new radius, $G M = g R^{2}$ )
$\Rightarrow \frac{- m g R}{4} + \frac{10^{4} R}{6} = - \frac{m g R^{2}}{2 r}$
$\Rightarrow \frac{- 10^{3} \times 10 R}{4} + \frac{10^{4} R}{6} = \frac{- 10^{3} \times 10 \times R^{2}}{2 r}$
$\Rightarrow r = 6 R$

JEE - 2024

PHXI08:GRAVITATION

359729 A satellite is orbiting with areal velocity $v_{a}$ in circular orbit. At what height from the surface of the earth, it is rotating, if the radius of earth is $R$ ?

1

\frac{4 v_{a}^{2}}{g R^{2}} - R

2

\frac{2 v_{a}^{2}}{g R^{2}} - R

3

\frac{v_{a}^{2}}{g R^{2}} - R

4

\frac{v_{a}^{2}}{2 g R^{2}} - R

Explanation:

Areal velocity $= \frac{L}{2 m}$
$v_{a} = \frac{1}{2 m} m v r$
$\begin{aligned} \Rightarrow v_{a} = \frac{v r}{2} = \frac{r}{2} \sqrt{\frac{G M}{r}} = \frac{r}{2} \sqrt{\frac{g R^{2}}{r}} = \frac{1}{2} \sqrt{g R^{2} r} \\ \Rightarrow \frac{4 v_{a}^{2}}{g R^{2}} = r = R + h \Rightarrow h = \frac{4 v_{a}^{2}}{g R^{2}} - R . \end{aligned}$

PHXI08:GRAVITATION

359730 The figure shows the variation of energy with the orbit radius of a body in circular planetary motion. Find the correct statement about the curves $A, B$ and $C$
supporting img

1

A

shows the kinetic energy,

B

the potential energy and

C

the total energy of the system

2

C

shows the total energy,

B

the kinetic energy and

A

the potential energy system

3

C

and

A

are kinetic and potential energies respectively and

B

is the total energy of the system

4

A

and

B

are the kinetic and potential energies respectively and

C

is the total energy of the system

Explanation:

Let $m$ is the mass of the body and $M$ is the mass of the earth. The orbital velocity is
$\begin{aligned} v_{o} = \sqrt{\frac{G M}{r}} \\ K . E = \frac{1}{2} m v_{o}^{2} = \frac{G M m}{2 r} \\ P . E = \frac{- G M m}{r} \\ T . E = \frac{- G M m}{2 r} \end{aligned}$

PHXI08:GRAVITATION

359731 Two satellites $P$ and $Q$ of same mass are revolving near the earth surface in the equitorial plane. The satellite $P$ moves in the direction of rotation of earth whereas $Q$ moves in the opposite direction. The ratio of their kinetic energies with respect to a frame attached to earth will be -

1

{(\frac{7437}{8363})}^{2}

2

{(\frac{8363}{7437})}^{2}

3

(\frac{7437}{8363})

4

(\frac{8363}{7437})

Explanation:

Linear velocity of earth,
$v_{e} = \frac{2 π R_{e}}{T_{e}} = \frac{6.28 \times 6.4 \times 10^{6}}{24 \times 3600} = 463 m / s$
Orbital velocity,
According to question, $v_{0} = \sqrt{R_{e} g} = 7.9 \times 10^{3} m / s$
According to question,
$v_{K P} = v_{0} - v_{e} = 7900 - 463 = 7437 m / s$
$v_{K Q} = v_{0} + v_{e} = 7900 + 463 = 8363 m / s$
$∴ \frac{E_{K Q}}{E_{K P}} = \frac{v_{K Q}^{2}}{v_{K P}^{2}} = {(\frac{8363}{7437})}^{2}$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: