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PHXI08:GRAVITATION

359783 A satellite (mass $m$ ) of moon revolves around it in a radius $n$ times the radius of moon ( $R$ ). Due to cosmic dust it experiences a resistance $F = α v^{2}$ . Find how long it will stay in the orbit before it falls onto the planets surface.

1

\frac{m}{α \sqrt{g R}} \sqrt{n}

2

\frac{m}{α} \frac{v_{i}}{v_{f}^{2}}

3

\frac{m}{α} \frac{(\sqrt{n} - 1)}{v}

4

(\sqrt{n} - 1) \frac{m}{α \sqrt{g R}}

Explanation:

Air resistance $F = - α v^{2}$ ,
where $v$ is the orbital velocity $v = \sqrt{\frac{G M}{r}}$
$r$ -distance of satellite from planet's centre.
$F = - \frac{α G M}{r}$
The work done by the resistance force
$d W = F d x = F v d t = \frac{G M α}{r} \sqrt{\frac{G M}{r}} d t$
$= \frac{{(G M)}^{3 / 2} α}{r^{3 / 2}} d t (1)$
The loss of energy of the satellite $= d E$ $\frac{d U}{d r} = \frac{d}{d r} (- \frac{G M m}{2 r}) = \frac{G M m}{2 r^{2}}$
$\Rightarrow d E = \frac{G M m}{2 r^{2}} d r (2)$
$\Rightarrow d E = \frac{G M m}{2 r^{2}} d r$
From work energy theorem the loss in $T . E$ of satellite is equal to work done by the resistance force.
$\begin{aligned} d E = - d W \\ \frac{G M m}{2 r^{2}} d r = \frac{- (G M)^{3 / 2}}{r^{3 / 2}} α d t \\ t = - \frac{m}{2 α \sqrt{G M}} \int_{n R}^{R} \frac{d r}{\sqrt{r}} = \frac{m \sqrt{R} (\sqrt{n} - 1)}{α \sqrt{G M}} \\ t = (\sqrt{n} - 1) \frac{m}{α \sqrt{g R}} \end{aligned}$

PHXI08:GRAVITATION

359784 A satellite is orbiting just above the surface of the earth with period $T$ . If $d$ is the density of the earth and $G$ is the universal constant of
gravitation, the quantity $\frac{3 π}{G d}$ represents:

1

T^{2}

2

T^{3}

3

\sqrt{T}

4

T

Explanation:

$T = \frac{2 π}{\sqrt{G M}} r^{3 / 2}$
$\Rightarrow T^{2} = \frac{4 π^{2} R^{3}}{G (\frac{4}{3} π R^{3} d)}$
Using $(r = R)$
$T^{2} = \frac{3 π}{G d}$
Correct option is (1).

NEET - 2023

PHXI08:GRAVITATION

359785 Two satellites of massess $m_{1}$ and $m_{2} (m_{1} > m_{2})$ are revolving round the earth in circular orbits of radius $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively. Which of the following statements is true regarding their speeds $v_{1}$ and $v_{2}$ ?

1

\frac{v_{1}}{r_{1}} = \frac{v_{2}}{r_{2}}

2

v_{1} = v_{2}

3

v_{1} < v_{2}

4

v_{1} > v_{2}

Explanation:

$v = \sqrt{\frac{G M}{r}}$ if $r_{1} > r_{2}$ then $v_{1} < v_{2}$
Orbital speed of satellite does not depend upon the mass of the satellite.

PHXI08:GRAVITATION

359786 A statellite is revolving in a circular orbit at a height $h$ above the surface of the earth of radius $R$ . The speed of the satellite in its orbit is one fourth the escape velocity from the surface of the earth. The relation between $h$ and $R$ is

1

h = 2 R

2

h = 3 R

3

h = 5 R

4

h = 7 R

Explanation:

Given, orbital speed $v = \frac{1}{4} v_{e}$
$\Rightarrow \sqrt{\frac{G M}{(R + h)}} = \frac{1}{4} \sqrt{\frac{2 G M}{R}}$
Squaring both sides,
$\begin{aligned} \frac{G M}{R + h} = \frac{1}{16} \frac{2 G M}{R} \\ \Rightarrow R + h = 8 R \\ \Rightarrow h = 7 R \end{aligned}$

MHTCET - 2018

PHXI08:GRAVITATION

359783 A satellite (mass $m$ ) of moon revolves around it in a radius $n$ times the radius of moon ( $R$ ). Due to cosmic dust it experiences a resistance $F = α v^{2}$ . Find how long it will stay in the orbit before it falls onto the planets surface.

1

\frac{m}{α \sqrt{g R}} \sqrt{n}

2

\frac{m}{α} \frac{v_{i}}{v_{f}^{2}}

3

\frac{m}{α} \frac{(\sqrt{n} - 1)}{v}

4

(\sqrt{n} - 1) \frac{m}{α \sqrt{g R}}

Explanation:

Air resistance $F = - α v^{2}$ ,
where $v$ is the orbital velocity $v = \sqrt{\frac{G M}{r}}$
$r$ -distance of satellite from planet's centre.
$F = - \frac{α G M}{r}$
The work done by the resistance force
$d W = F d x = F v d t = \frac{G M α}{r} \sqrt{\frac{G M}{r}} d t$
$= \frac{{(G M)}^{3 / 2} α}{r^{3 / 2}} d t (1)$
The loss of energy of the satellite $= d E$ $\frac{d U}{d r} = \frac{d}{d r} (- \frac{G M m}{2 r}) = \frac{G M m}{2 r^{2}}$
$\Rightarrow d E = \frac{G M m}{2 r^{2}} d r (2)$
$\Rightarrow d E = \frac{G M m}{2 r^{2}} d r$
From work energy theorem the loss in $T . E$ of satellite is equal to work done by the resistance force.
$\begin{aligned} d E = - d W \\ \frac{G M m}{2 r^{2}} d r = \frac{- (G M)^{3 / 2}}{r^{3 / 2}} α d t \\ t = - \frac{m}{2 α \sqrt{G M}} \int_{n R}^{R} \frac{d r}{\sqrt{r}} = \frac{m \sqrt{R} (\sqrt{n} - 1)}{α \sqrt{G M}} \\ t = (\sqrt{n} - 1) \frac{m}{α \sqrt{g R}} \end{aligned}$

PHXI08:GRAVITATION

359784 A satellite is orbiting just above the surface of the earth with period $T$ . If $d$ is the density of the earth and $G$ is the universal constant of
gravitation, the quantity $\frac{3 π}{G d}$ represents:

1

T^{2}

2

T^{3}

3

\sqrt{T}

4

T

Explanation:

$T = \frac{2 π}{\sqrt{G M}} r^{3 / 2}$
$\Rightarrow T^{2} = \frac{4 π^{2} R^{3}}{G (\frac{4}{3} π R^{3} d)}$
Using $(r = R)$
$T^{2} = \frac{3 π}{G d}$
Correct option is (1).

NEET - 2023

PHXI08:GRAVITATION

359785 Two satellites of massess $m_{1}$ and $m_{2} (m_{1} > m_{2})$ are revolving round the earth in circular orbits of radius $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively. Which of the following statements is true regarding their speeds $v_{1}$ and $v_{2}$ ?

1

\frac{v_{1}}{r_{1}} = \frac{v_{2}}{r_{2}}

2

v_{1} = v_{2}

3

v_{1} < v_{2}

4

v_{1} > v_{2}

Explanation:

$v = \sqrt{\frac{G M}{r}}$ if $r_{1} > r_{2}$ then $v_{1} < v_{2}$
Orbital speed of satellite does not depend upon the mass of the satellite.

PHXI08:GRAVITATION

359786 A statellite is revolving in a circular orbit at a height $h$ above the surface of the earth of radius $R$ . The speed of the satellite in its orbit is one fourth the escape velocity from the surface of the earth. The relation between $h$ and $R$ is

1

h = 2 R

2

h = 3 R

3

h = 5 R

4

h = 7 R

Explanation:

Given, orbital speed $v = \frac{1}{4} v_{e}$
$\Rightarrow \sqrt{\frac{G M}{(R + h)}} = \frac{1}{4} \sqrt{\frac{2 G M}{R}}$
Squaring both sides,
$\begin{aligned} \frac{G M}{R + h} = \frac{1}{16} \frac{2 G M}{R} \\ \Rightarrow R + h = 8 R \\ \Rightarrow h = 7 R \end{aligned}$

MHTCET - 2018

PHXI08:GRAVITATION

359783 A satellite (mass $m$ ) of moon revolves around it in a radius $n$ times the radius of moon ( $R$ ). Due to cosmic dust it experiences a resistance $F = α v^{2}$ . Find how long it will stay in the orbit before it falls onto the planets surface.

1

\frac{m}{α \sqrt{g R}} \sqrt{n}

2

\frac{m}{α} \frac{v_{i}}{v_{f}^{2}}

3

\frac{m}{α} \frac{(\sqrt{n} - 1)}{v}

4

(\sqrt{n} - 1) \frac{m}{α \sqrt{g R}}

Explanation:

Air resistance $F = - α v^{2}$ ,
where $v$ is the orbital velocity $v = \sqrt{\frac{G M}{r}}$
$r$ -distance of satellite from planet's centre.
$F = - \frac{α G M}{r}$
The work done by the resistance force
$d W = F d x = F v d t = \frac{G M α}{r} \sqrt{\frac{G M}{r}} d t$
$= \frac{{(G M)}^{3 / 2} α}{r^{3 / 2}} d t (1)$
The loss of energy of the satellite $= d E$ $\frac{d U}{d r} = \frac{d}{d r} (- \frac{G M m}{2 r}) = \frac{G M m}{2 r^{2}}$
$\Rightarrow d E = \frac{G M m}{2 r^{2}} d r (2)$
$\Rightarrow d E = \frac{G M m}{2 r^{2}} d r$
From work energy theorem the loss in $T . E$ of satellite is equal to work done by the resistance force.
$\begin{aligned} d E = - d W \\ \frac{G M m}{2 r^{2}} d r = \frac{- (G M)^{3 / 2}}{r^{3 / 2}} α d t \\ t = - \frac{m}{2 α \sqrt{G M}} \int_{n R}^{R} \frac{d r}{\sqrt{r}} = \frac{m \sqrt{R} (\sqrt{n} - 1)}{α \sqrt{G M}} \\ t = (\sqrt{n} - 1) \frac{m}{α \sqrt{g R}} \end{aligned}$

PHXI08:GRAVITATION

359784 A satellite is orbiting just above the surface of the earth with period $T$ . If $d$ is the density of the earth and $G$ is the universal constant of
gravitation, the quantity $\frac{3 π}{G d}$ represents:

1

T^{2}

2

T^{3}

3

\sqrt{T}

4

T

Explanation:

$T = \frac{2 π}{\sqrt{G M}} r^{3 / 2}$
$\Rightarrow T^{2} = \frac{4 π^{2} R^{3}}{G (\frac{4}{3} π R^{3} d)}$
Using $(r = R)$
$T^{2} = \frac{3 π}{G d}$
Correct option is (1).

NEET - 2023

PHXI08:GRAVITATION

359785 Two satellites of massess $m_{1}$ and $m_{2} (m_{1} > m_{2})$ are revolving round the earth in circular orbits of radius $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively. Which of the following statements is true regarding their speeds $v_{1}$ and $v_{2}$ ?

1

\frac{v_{1}}{r_{1}} = \frac{v_{2}}{r_{2}}

2

v_{1} = v_{2}

3

v_{1} < v_{2}

4

v_{1} > v_{2}

Explanation:

$v = \sqrt{\frac{G M}{r}}$ if $r_{1} > r_{2}$ then $v_{1} < v_{2}$
Orbital speed of satellite does not depend upon the mass of the satellite.

PHXI08:GRAVITATION

359786 A statellite is revolving in a circular orbit at a height $h$ above the surface of the earth of radius $R$ . The speed of the satellite in its orbit is one fourth the escape velocity from the surface of the earth. The relation between $h$ and $R$ is

1

h = 2 R

2

h = 3 R

3

h = 5 R

4

h = 7 R

Explanation:

Given, orbital speed $v = \frac{1}{4} v_{e}$
$\Rightarrow \sqrt{\frac{G M}{(R + h)}} = \frac{1}{4} \sqrt{\frac{2 G M}{R}}$
Squaring both sides,
$\begin{aligned} \frac{G M}{R + h} = \frac{1}{16} \frac{2 G M}{R} \\ \Rightarrow R + h = 8 R \\ \Rightarrow h = 7 R \end{aligned}$

MHTCET - 2018

PHXI08:GRAVITATION

359783 A satellite (mass $m$ ) of moon revolves around it in a radius $n$ times the radius of moon ( $R$ ). Due to cosmic dust it experiences a resistance $F = α v^{2}$ . Find how long it will stay in the orbit before it falls onto the planets surface.

1

\frac{m}{α \sqrt{g R}} \sqrt{n}

2

\frac{m}{α} \frac{v_{i}}{v_{f}^{2}}

3

\frac{m}{α} \frac{(\sqrt{n} - 1)}{v}

4

(\sqrt{n} - 1) \frac{m}{α \sqrt{g R}}

Explanation:

Air resistance $F = - α v^{2}$ ,
where $v$ is the orbital velocity $v = \sqrt{\frac{G M}{r}}$
$r$ -distance of satellite from planet's centre.
$F = - \frac{α G M}{r}$
The work done by the resistance force
$d W = F d x = F v d t = \frac{G M α}{r} \sqrt{\frac{G M}{r}} d t$
$= \frac{{(G M)}^{3 / 2} α}{r^{3 / 2}} d t (1)$
The loss of energy of the satellite $= d E$ $\frac{d U}{d r} = \frac{d}{d r} (- \frac{G M m}{2 r}) = \frac{G M m}{2 r^{2}}$
$\Rightarrow d E = \frac{G M m}{2 r^{2}} d r (2)$
$\Rightarrow d E = \frac{G M m}{2 r^{2}} d r$
From work energy theorem the loss in $T . E$ of satellite is equal to work done by the resistance force.
$\begin{aligned} d E = - d W \\ \frac{G M m}{2 r^{2}} d r = \frac{- (G M)^{3 / 2}}{r^{3 / 2}} α d t \\ t = - \frac{m}{2 α \sqrt{G M}} \int_{n R}^{R} \frac{d r}{\sqrt{r}} = \frac{m \sqrt{R} (\sqrt{n} - 1)}{α \sqrt{G M}} \\ t = (\sqrt{n} - 1) \frac{m}{α \sqrt{g R}} \end{aligned}$

PHXI08:GRAVITATION

359784 A satellite is orbiting just above the surface of the earth with period $T$ . If $d$ is the density of the earth and $G$ is the universal constant of
gravitation, the quantity $\frac{3 π}{G d}$ represents:

1

T^{2}

2

T^{3}

3

\sqrt{T}

4

T

Explanation:

$T = \frac{2 π}{\sqrt{G M}} r^{3 / 2}$
$\Rightarrow T^{2} = \frac{4 π^{2} R^{3}}{G (\frac{4}{3} π R^{3} d)}$
Using $(r = R)$
$T^{2} = \frac{3 π}{G d}$
Correct option is (1).

NEET - 2023

PHXI08:GRAVITATION

359785 Two satellites of massess $m_{1}$ and $m_{2} (m_{1} > m_{2})$ are revolving round the earth in circular orbits of radius $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively. Which of the following statements is true regarding their speeds $v_{1}$ and $v_{2}$ ?

1

\frac{v_{1}}{r_{1}} = \frac{v_{2}}{r_{2}}

2

v_{1} = v_{2}

3

v_{1} < v_{2}

4

v_{1} > v_{2}

Explanation:

$v = \sqrt{\frac{G M}{r}}$ if $r_{1} > r_{2}$ then $v_{1} < v_{2}$
Orbital speed of satellite does not depend upon the mass of the satellite.

PHXI08:GRAVITATION

359786 A statellite is revolving in a circular orbit at a height $h$ above the surface of the earth of radius $R$ . The speed of the satellite in its orbit is one fourth the escape velocity from the surface of the earth. The relation between $h$ and $R$ is

1

h = 2 R

2

h = 3 R

3

h = 5 R

4

h = 7 R

Explanation:

Given, orbital speed $v = \frac{1}{4} v_{e}$
$\Rightarrow \sqrt{\frac{G M}{(R + h)}} = \frac{1}{4} \sqrt{\frac{2 G M}{R}}$
Squaring both sides,
$\begin{aligned} \frac{G M}{R + h} = \frac{1}{16} \frac{2 G M}{R} \\ \Rightarrow R + h = 8 R \\ \Rightarrow h = 7 R \end{aligned}$

MHTCET - 2018