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PHXI08:GRAVITATION

359753 Two satellites of masses $m$ and $3 m$ revolve around the earth in circular orbits of radii $r$ and $3 r$ respectively. The ratio of orbital speeds of the satellites respectively is

1

9 : 1

2

\sqrt{3} : 1

3

3 : 1

4

1 : 1

Explanation:

Orbital speed of satellite is, $v = \sqrt{\frac{G M}{r}}$
$∴ v \propto \frac{1}{\sqrt{r}}$
So, as $r$ increases, $v$ decreases
Given : $r_{1} = r, r_{2} = 3 r$
$\Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{r_{2}}{r_{1}}} = \sqrt{\frac{3 r}{r}} = \sqrt{3} : 1$
So, correct option is (2).

JEE - 2023

PHXI08:GRAVITATION

359754 A satellite is moving on a circular path of radius $r$ around the earth has a time period $T$ . If its radius slightly increased by $Δ r$ , the change in its time period is:

1

\frac{3}{2} (\frac{T}{r}) Δ r

2

(\frac{T}{r}) Δ r

3

\frac{3}{2} (\frac{T}{r}) Δ T

4 None of these

Explanation:

$T^{2} \propto r^{3}$
$\begin{aligned} ∴ 2 Δ \frac{T}{T} = 3 \frac{Δ r}{r} \\ ∴ Δ T = \frac{3}{2} (\frac{Δ r}{r}) \times T \end{aligned}$

PHXI08:GRAVITATION

359755 The time period ' $T$ ' of a satellite is related to the density $(ρ)$ of the planet which is orbiting close around the planet as

1

T \propto ρ^{\frac{- 1}{2}}

2

T \propto ρ

3

T \propto ρ^{\frac{1}{2}}

4

T \propto ρ^{\frac{- 3}{2}}

Explanation:

Time period of satellite is given by,
$T = 2 π \sqrt{\frac{r^{3}}{G M}}$
Where ' $r$ ' is the radius of the orbit and ' $M$ ' is the mass of the planet,
As the satellite is revolving very close to the planet, so $r = R$ .
Also, $M = ρ \times V = ρ \times \frac{4}{3} π R^{3}$
$T = 2 π \sqrt{\frac{R^{3}}{G \times \frac{4 π}{3} ρ R^{3}}} = \sqrt{\frac{3 π}{ρ G}} i.e. T \propto ρ^{- \frac{1}{2}}$

MHTCET - 2022

PHXI08:GRAVITATION

359756 Assertion :
The speed of revolution of an artificial satellite revolving very near the earth is $8 {kms}^{- 1}$
Reason :
Orbital velocity of a satellite, become independent of height of satellite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Orbital velocity
$v_{0} = \sqrt{R_{E} g} = \sqrt{6400 \times 10^{3} \times 10}$
$\Rightarrow v_{0} = 8 \times 10^{3} m / s = 8 k m / s$
Here $(R_{E} + h) \approx R_{E} ($ since $h << R_{E})$ . Assertion says "very near to surface of earth". The orbital velocity of a satellite depends on the mass of the Earth and the radius of the satellite's orbit, but it is largely independent of the height $h$ of the satellite above the Earth's surface. (Since $h$ is small compared to $R_{E}$ ). So correct option is (1).

PHXI08:GRAVITATION

359753 Two satellites of masses $m$ and $3 m$ revolve around the earth in circular orbits of radii $r$ and $3 r$ respectively. The ratio of orbital speeds of the satellites respectively is

1

9 : 1

2

\sqrt{3} : 1

3

3 : 1

4

1 : 1

Explanation:

Orbital speed of satellite is, $v = \sqrt{\frac{G M}{r}}$
$∴ v \propto \frac{1}{\sqrt{r}}$
So, as $r$ increases, $v$ decreases
Given : $r_{1} = r, r_{2} = 3 r$
$\Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{r_{2}}{r_{1}}} = \sqrt{\frac{3 r}{r}} = \sqrt{3} : 1$
So, correct option is (2).

JEE - 2023

PHXI08:GRAVITATION

359754 A satellite is moving on a circular path of radius $r$ around the earth has a time period $T$ . If its radius slightly increased by $Δ r$ , the change in its time period is:

1

\frac{3}{2} (\frac{T}{r}) Δ r

2

(\frac{T}{r}) Δ r

3

\frac{3}{2} (\frac{T}{r}) Δ T

4 None of these

Explanation:

$T^{2} \propto r^{3}$
$\begin{aligned} ∴ 2 Δ \frac{T}{T} = 3 \frac{Δ r}{r} \\ ∴ Δ T = \frac{3}{2} (\frac{Δ r}{r}) \times T \end{aligned}$

PHXI08:GRAVITATION

359755 The time period ' $T$ ' of a satellite is related to the density $(ρ)$ of the planet which is orbiting close around the planet as

1

T \propto ρ^{\frac{- 1}{2}}

2

T \propto ρ

3

T \propto ρ^{\frac{1}{2}}

4

T \propto ρ^{\frac{- 3}{2}}

Explanation:

Time period of satellite is given by,
$T = 2 π \sqrt{\frac{r^{3}}{G M}}$
Where ' $r$ ' is the radius of the orbit and ' $M$ ' is the mass of the planet,
As the satellite is revolving very close to the planet, so $r = R$ .
Also, $M = ρ \times V = ρ \times \frac{4}{3} π R^{3}$
$T = 2 π \sqrt{\frac{R^{3}}{G \times \frac{4 π}{3} ρ R^{3}}} = \sqrt{\frac{3 π}{ρ G}} i.e. T \propto ρ^{- \frac{1}{2}}$

MHTCET - 2022

PHXI08:GRAVITATION

359756 Assertion :
The speed of revolution of an artificial satellite revolving very near the earth is $8 {kms}^{- 1}$
Reason :
Orbital velocity of a satellite, become independent of height of satellite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Orbital velocity
$v_{0} = \sqrt{R_{E} g} = \sqrt{6400 \times 10^{3} \times 10}$
$\Rightarrow v_{0} = 8 \times 10^{3} m / s = 8 k m / s$
Here $(R_{E} + h) \approx R_{E} ($ since $h << R_{E})$ . Assertion says "very near to surface of earth". The orbital velocity of a satellite depends on the mass of the Earth and the radius of the satellite's orbit, but it is largely independent of the height $h$ of the satellite above the Earth's surface. (Since $h$ is small compared to $R_{E}$ ). So correct option is (1).

PHXI08:GRAVITATION

359753 Two satellites of masses $m$ and $3 m$ revolve around the earth in circular orbits of radii $r$ and $3 r$ respectively. The ratio of orbital speeds of the satellites respectively is

1

9 : 1

2

\sqrt{3} : 1

3

3 : 1

4

1 : 1

Explanation:

Orbital speed of satellite is, $v = \sqrt{\frac{G M}{r}}$
$∴ v \propto \frac{1}{\sqrt{r}}$
So, as $r$ increases, $v$ decreases
Given : $r_{1} = r, r_{2} = 3 r$
$\Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{r_{2}}{r_{1}}} = \sqrt{\frac{3 r}{r}} = \sqrt{3} : 1$
So, correct option is (2).

JEE - 2023

PHXI08:GRAVITATION

359754 A satellite is moving on a circular path of radius $r$ around the earth has a time period $T$ . If its radius slightly increased by $Δ r$ , the change in its time period is:

1

\frac{3}{2} (\frac{T}{r}) Δ r

2

(\frac{T}{r}) Δ r

3

\frac{3}{2} (\frac{T}{r}) Δ T

4 None of these

Explanation:

$T^{2} \propto r^{3}$
$\begin{aligned} ∴ 2 Δ \frac{T}{T} = 3 \frac{Δ r}{r} \\ ∴ Δ T = \frac{3}{2} (\frac{Δ r}{r}) \times T \end{aligned}$

PHXI08:GRAVITATION

359755 The time period ' $T$ ' of a satellite is related to the density $(ρ)$ of the planet which is orbiting close around the planet as

1

T \propto ρ^{\frac{- 1}{2}}

2

T \propto ρ

3

T \propto ρ^{\frac{1}{2}}

4

T \propto ρ^{\frac{- 3}{2}}

Explanation:

Time period of satellite is given by,
$T = 2 π \sqrt{\frac{r^{3}}{G M}}$
Where ' $r$ ' is the radius of the orbit and ' $M$ ' is the mass of the planet,
As the satellite is revolving very close to the planet, so $r = R$ .
Also, $M = ρ \times V = ρ \times \frac{4}{3} π R^{3}$
$T = 2 π \sqrt{\frac{R^{3}}{G \times \frac{4 π}{3} ρ R^{3}}} = \sqrt{\frac{3 π}{ρ G}} i.e. T \propto ρ^{- \frac{1}{2}}$

MHTCET - 2022

PHXI08:GRAVITATION

359756 Assertion :
The speed of revolution of an artificial satellite revolving very near the earth is $8 {kms}^{- 1}$
Reason :
Orbital velocity of a satellite, become independent of height of satellite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Orbital velocity
$v_{0} = \sqrt{R_{E} g} = \sqrt{6400 \times 10^{3} \times 10}$
$\Rightarrow v_{0} = 8 \times 10^{3} m / s = 8 k m / s$
Here $(R_{E} + h) \approx R_{E} ($ since $h << R_{E})$ . Assertion says "very near to surface of earth". The orbital velocity of a satellite depends on the mass of the Earth and the radius of the satellite's orbit, but it is largely independent of the height $h$ of the satellite above the Earth's surface. (Since $h$ is small compared to $R_{E}$ ). So correct option is (1).

PHXI08:GRAVITATION

359753 Two satellites of masses $m$ and $3 m$ revolve around the earth in circular orbits of radii $r$ and $3 r$ respectively. The ratio of orbital speeds of the satellites respectively is

1

9 : 1

2

\sqrt{3} : 1

3

3 : 1

4

1 : 1

Explanation:

Orbital speed of satellite is, $v = \sqrt{\frac{G M}{r}}$
$∴ v \propto \frac{1}{\sqrt{r}}$
So, as $r$ increases, $v$ decreases
Given : $r_{1} = r, r_{2} = 3 r$
$\Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{r_{2}}{r_{1}}} = \sqrt{\frac{3 r}{r}} = \sqrt{3} : 1$
So, correct option is (2).

JEE - 2023

PHXI08:GRAVITATION

359754 A satellite is moving on a circular path of radius $r$ around the earth has a time period $T$ . If its radius slightly increased by $Δ r$ , the change in its time period is:

1

\frac{3}{2} (\frac{T}{r}) Δ r

2

(\frac{T}{r}) Δ r

3

\frac{3}{2} (\frac{T}{r}) Δ T

4 None of these

Explanation:

$T^{2} \propto r^{3}$
$\begin{aligned} ∴ 2 Δ \frac{T}{T} = 3 \frac{Δ r}{r} \\ ∴ Δ T = \frac{3}{2} (\frac{Δ r}{r}) \times T \end{aligned}$

PHXI08:GRAVITATION

359755 The time period ' $T$ ' of a satellite is related to the density $(ρ)$ of the planet which is orbiting close around the planet as

1

T \propto ρ^{\frac{- 1}{2}}

2

T \propto ρ

3

T \propto ρ^{\frac{1}{2}}

4

T \propto ρ^{\frac{- 3}{2}}

Explanation:

Time period of satellite is given by,
$T = 2 π \sqrt{\frac{r^{3}}{G M}}$
Where ' $r$ ' is the radius of the orbit and ' $M$ ' is the mass of the planet,
As the satellite is revolving very close to the planet, so $r = R$ .
Also, $M = ρ \times V = ρ \times \frac{4}{3} π R^{3}$
$T = 2 π \sqrt{\frac{R^{3}}{G \times \frac{4 π}{3} ρ R^{3}}} = \sqrt{\frac{3 π}{ρ G}} i.e. T \propto ρ^{- \frac{1}{2}}$

MHTCET - 2022

PHXI08:GRAVITATION

359756 Assertion :
The speed of revolution of an artificial satellite revolving very near the earth is $8 {kms}^{- 1}$
Reason :
Orbital velocity of a satellite, become independent of height of satellite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Orbital velocity
$v_{0} = \sqrt{R_{E} g} = \sqrt{6400 \times 10^{3} \times 10}$
$\Rightarrow v_{0} = 8 \times 10^{3} m / s = 8 k m / s$
Here $(R_{E} + h) \approx R_{E} ($ since $h << R_{E})$ . Assertion says "very near to surface of earth". The orbital velocity of a satellite depends on the mass of the Earth and the radius of the satellite's orbit, but it is largely independent of the height $h$ of the satellite above the Earth's surface. (Since $h$ is small compared to $R_{E}$ ). So correct option is (1).

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