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PHXI08:GRAVITATION

359749 Statement A :
Electrical force is experienced by charged particles only.
Statement B :
There is no gravitational force on the satellite.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

PHXI08:GRAVITATION

359750 Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

1

R^{(\frac{n - 1}{2})}

2

R^{(\frac{n - 2}{2})}

3

R^{(\frac{n + 1}{2})}

4

R^{n}

Explanation:

$m ω^{2} R = \frac{G M m}{R^{n}} \Rightarrow m (\frac{4 π^{2}}{T^{2}}) R \propto \frac{1}{R^{n}} \Rightarrow T^{2} \propto R^{n + 1}$
$∴ T \propto R^{(\frac{n + 1}{2})}$

PHXI08:GRAVITATION

359751 An earth satellite is moved from one stable circular orbit to another larger and stable circular orbit. The following quantities increase for the satellite as a result of this change

1 Gravitational potential energy

2 Angular vleocity

3 Linear orbital velocity

4 Centripetal acceleration

Explanation:

When a satellite is in stable orbit then its time period is
$\begin{matrix} T = 2 π \sqrt{\frac{r^{3}}{G M}} = \frac{2 π}{ω} \\ ω = \sqrt{\frac{G M}{r^{3}}} \Rightarrow ω α \sqrt{\frac{1}{r^{3}}} \\ v_{0} = \sqrt{\frac{G M}{r}} \Rightarrow v_{0} α \sqrt{\frac{1}{r}} \\ U = \frac{- G M m}{r} As r increases U also increases \end{matrix}$
The centripetal acceleration is
$a_{r} = \frac{v_{0}^{2}}{r} = \frac{G M}{r^{2}} \Rightarrow a_{r} α \frac{1}{r}$

PHXI08:GRAVITATION

359752 Two satellites $A$ and $B$ go round a planet in circular orbits having radii $4 R$ and $R$ respectively. If the speed of $A$ is $3 v$ , the speed of $B$ will be

1

12 v

2

\frac{4}{3} v

3

6 v

4

3 v

Explanation:

Orbital speed is given by $v_{0} = \sqrt{\frac{G M}{R}}$ where $r =$ radius of orbit
$v_{orbital} = \sqrt{\frac{G M}{R}}$
$v_{A} = 3 v = \sqrt{\frac{G M}{4 R}}$ $(1)$
and $v_{B} = \sqrt{\frac{G M}{R}}$ $(2)$
From eq (1) and (2)
$\frac{3 v}{v_{B}} = \frac{1}{2} ∴ v_{B} = 6 v$

JEE - 2024

PHXI08:GRAVITATION

359749 Statement A :
Electrical force is experienced by charged particles only.
Statement B :
There is no gravitational force on the satellite.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

PHXI08:GRAVITATION

359750 Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

1

R^{(\frac{n - 1}{2})}

2

R^{(\frac{n - 2}{2})}

3

R^{(\frac{n + 1}{2})}

4

R^{n}

Explanation:

$m ω^{2} R = \frac{G M m}{R^{n}} \Rightarrow m (\frac{4 π^{2}}{T^{2}}) R \propto \frac{1}{R^{n}} \Rightarrow T^{2} \propto R^{n + 1}$
$∴ T \propto R^{(\frac{n + 1}{2})}$

PHXI08:GRAVITATION

359751 An earth satellite is moved from one stable circular orbit to another larger and stable circular orbit. The following quantities increase for the satellite as a result of this change

1 Gravitational potential energy

2 Angular vleocity

3 Linear orbital velocity

4 Centripetal acceleration

Explanation:

When a satellite is in stable orbit then its time period is
$\begin{matrix} T = 2 π \sqrt{\frac{r^{3}}{G M}} = \frac{2 π}{ω} \\ ω = \sqrt{\frac{G M}{r^{3}}} \Rightarrow ω α \sqrt{\frac{1}{r^{3}}} \\ v_{0} = \sqrt{\frac{G M}{r}} \Rightarrow v_{0} α \sqrt{\frac{1}{r}} \\ U = \frac{- G M m}{r} As r increases U also increases \end{matrix}$
The centripetal acceleration is
$a_{r} = \frac{v_{0}^{2}}{r} = \frac{G M}{r^{2}} \Rightarrow a_{r} α \frac{1}{r}$

PHXI08:GRAVITATION

359752 Two satellites $A$ and $B$ go round a planet in circular orbits having radii $4 R$ and $R$ respectively. If the speed of $A$ is $3 v$ , the speed of $B$ will be

1

12 v

2

\frac{4}{3} v

3

6 v

4

3 v

Explanation:

Orbital speed is given by $v_{0} = \sqrt{\frac{G M}{R}}$ where $r =$ radius of orbit
$v_{orbital} = \sqrt{\frac{G M}{R}}$
$v_{A} = 3 v = \sqrt{\frac{G M}{4 R}}$ $(1)$
and $v_{B} = \sqrt{\frac{G M}{R}}$ $(2)$
From eq (1) and (2)
$\frac{3 v}{v_{B}} = \frac{1}{2} ∴ v_{B} = 6 v$

JEE - 2024

PHXI08:GRAVITATION

359749 Statement A :
Electrical force is experienced by charged particles only.
Statement B :
There is no gravitational force on the satellite.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

PHXI08:GRAVITATION

359750 Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

1

R^{(\frac{n - 1}{2})}

2

R^{(\frac{n - 2}{2})}

3

R^{(\frac{n + 1}{2})}

4

R^{n}

Explanation:

$m ω^{2} R = \frac{G M m}{R^{n}} \Rightarrow m (\frac{4 π^{2}}{T^{2}}) R \propto \frac{1}{R^{n}} \Rightarrow T^{2} \propto R^{n + 1}$
$∴ T \propto R^{(\frac{n + 1}{2})}$

PHXI08:GRAVITATION

359751 An earth satellite is moved from one stable circular orbit to another larger and stable circular orbit. The following quantities increase for the satellite as a result of this change

1 Gravitational potential energy

2 Angular vleocity

3 Linear orbital velocity

4 Centripetal acceleration

Explanation:

When a satellite is in stable orbit then its time period is
$\begin{matrix} T = 2 π \sqrt{\frac{r^{3}}{G M}} = \frac{2 π}{ω} \\ ω = \sqrt{\frac{G M}{r^{3}}} \Rightarrow ω α \sqrt{\frac{1}{r^{3}}} \\ v_{0} = \sqrt{\frac{G M}{r}} \Rightarrow v_{0} α \sqrt{\frac{1}{r}} \\ U = \frac{- G M m}{r} As r increases U also increases \end{matrix}$
The centripetal acceleration is
$a_{r} = \frac{v_{0}^{2}}{r} = \frac{G M}{r^{2}} \Rightarrow a_{r} α \frac{1}{r}$

PHXI08:GRAVITATION

359752 Two satellites $A$ and $B$ go round a planet in circular orbits having radii $4 R$ and $R$ respectively. If the speed of $A$ is $3 v$ , the speed of $B$ will be

1

12 v

2

\frac{4}{3} v

3

6 v

4

3 v

Explanation:

Orbital speed is given by $v_{0} = \sqrt{\frac{G M}{R}}$ where $r =$ radius of orbit
$v_{orbital} = \sqrt{\frac{G M}{R}}$
$v_{A} = 3 v = \sqrt{\frac{G M}{4 R}}$ $(1)$
and $v_{B} = \sqrt{\frac{G M}{R}}$ $(2)$
From eq (1) and (2)
$\frac{3 v}{v_{B}} = \frac{1}{2} ∴ v_{B} = 6 v$

JEE - 2024

PHXI08:GRAVITATION

359749 Statement A :
Electrical force is experienced by charged particles only.
Statement B :
There is no gravitational force on the satellite.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

PHXI08:GRAVITATION

359750 Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

1

R^{(\frac{n - 1}{2})}

2

R^{(\frac{n - 2}{2})}

3

R^{(\frac{n + 1}{2})}

4

R^{n}

Explanation:

$m ω^{2} R = \frac{G M m}{R^{n}} \Rightarrow m (\frac{4 π^{2}}{T^{2}}) R \propto \frac{1}{R^{n}} \Rightarrow T^{2} \propto R^{n + 1}$
$∴ T \propto R^{(\frac{n + 1}{2})}$

PHXI08:GRAVITATION

359751 An earth satellite is moved from one stable circular orbit to another larger and stable circular orbit. The following quantities increase for the satellite as a result of this change

1 Gravitational potential energy

2 Angular vleocity

3 Linear orbital velocity

4 Centripetal acceleration

Explanation:

When a satellite is in stable orbit then its time period is
$\begin{matrix} T = 2 π \sqrt{\frac{r^{3}}{G M}} = \frac{2 π}{ω} \\ ω = \sqrt{\frac{G M}{r^{3}}} \Rightarrow ω α \sqrt{\frac{1}{r^{3}}} \\ v_{0} = \sqrt{\frac{G M}{r}} \Rightarrow v_{0} α \sqrt{\frac{1}{r}} \\ U = \frac{- G M m}{r} As r increases U also increases \end{matrix}$
The centripetal acceleration is
$a_{r} = \frac{v_{0}^{2}}{r} = \frac{G M}{r^{2}} \Rightarrow a_{r} α \frac{1}{r}$

PHXI08:GRAVITATION

359752 Two satellites $A$ and $B$ go round a planet in circular orbits having radii $4 R$ and $R$ respectively. If the speed of $A$ is $3 v$ , the speed of $B$ will be

1

12 v

2

\frac{4}{3} v

3

6 v

4

3 v

Explanation:

Orbital speed is given by $v_{0} = \sqrt{\frac{G M}{R}}$ where $r =$ radius of orbit
$v_{orbital} = \sqrt{\frac{G M}{R}}$
$v_{A} = 3 v = \sqrt{\frac{G M}{4 R}}$ $(1)$
and $v_{B} = \sqrt{\frac{G M}{R}}$ $(2)$
From eq (1) and (2)
$\frac{3 v}{v_{B}} = \frac{1}{2} ∴ v_{B} = 6 v$

JEE - 2024

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