QBManager

Gravitation

138696 If the radius of a planet is $R$ and its density is $ρ$ , the escape velocity from its surface will be

1

v_{e} \propto ρ R

2

v_{e} \propto \sqrt{ρ} R

3

v_{e} \propto \sqrt{\frac{ρ}{R}}

4

v_{e} \propto \frac{1}{\sqrt{ρ R}}

Explanation:

B We know that,
Escape velocity $(v_{e}) = \sqrt{\frac{2 GM}{R}}$
$v_{e} = \sqrt{\frac{2 G}{R} \cdot \frac{4}{3} π R^{3} ρ} (\begin{aligned} ∵ M & = V ρ \\ = \frac{4}{3} π R^{3} ρ \end{aligned})$
$v_{e} = R \sqrt{\frac{8 G π ρ}{3}}$
$v_{e} \propto R \sqrt{ρ}$
Hence, option (b) is correct.

CG PET- 2005

Gravitation

138697 The escape velocity from the earth's surface is $11.2 Km / s$ . If a planet has a radius twice that of the earth and on which the acceleration due to gravity is twice that on the earth, then the escape velocity on this planet will be

1

11.2 km / s

2

5.6 km / s

3

22.4 km / s

4

33.6 km / s

Explanation:

C Given that,
$v_{e} = 11.2 km / s$
$R_{P} = 2 R_{e}$
$g_{P} = 2 g_{e}$
$v_{e} = \sqrt{2 gR}$
Then, $\frac{v_{P}}{v_{e}} = \frac{\sqrt{2 g_{p} R_{p}}}{\sqrt{2 g_{e} R_{e}}}$
$\frac{v_{P}}{11.2} = \frac{\sqrt{2 \times 2 g_{e} \times 2 R_{e}}}{2 g_{e} R_{e}}$
$v_{P} = 2 \times 11.2 = 22.4 km / s$

CG PET-2004

Gravitation

138698 A satellite revolves very near to the earth surface. Its speed should be around

1

5 km / s

2

8 km / s

3

2 km / s

4

11 km / s

Explanation:

B Since, orbital speed of satellite near to the earth surface is approximately $8 km / s$ .
We know that,
$orbital velocity (v) = \sqrt{\frac{GM}{R}}$
$= \sqrt{\frac{6.67 \times 10^{- 11} \times 6 \times 10^{24}}{6400 \times 1000}}$
$= 7.9 \times 10^{3} m / s$
$\approx 8 km / s$

Manipal UGET-2019

Gravitation

138699 For a given density of a planet, the orbital speed of satellite near the surface of the planet of radius $R$ is proportional to

1

R^{1 / 2}

2

R^{3 / 2}

3

R^{- 1 / 2}

4

R^{0}

Explanation:

D From the Kepler's third law -
$\frac{T^{2}}{R^{3}} = \frac{4 π^{2}}{GM}$
when $r$ is radius, semi-major axis of the elliptical near the planet surface.
$M = ρ V$
$= ρ \times \frac{4}{3} π R^{3}$
$\frac{T^{2}}{R^{3}} = \frac{4 π^{2}}{G . ρ \times \frac{4}{3} π R^{3}}$
$T = \sqrt{\frac{3 π}{G ρ}} \times R^{0}$
The planet of radius $R$ is proportional to $R^{0}$ .

Manipal UGET-2015

Gravitation

138696 If the radius of a planet is $R$ and its density is $ρ$ , the escape velocity from its surface will be

1

v_{e} \propto ρ R

2

v_{e} \propto \sqrt{ρ} R

3

v_{e} \propto \sqrt{\frac{ρ}{R}}

4

v_{e} \propto \frac{1}{\sqrt{ρ R}}

Explanation:

B We know that,
Escape velocity $(v_{e}) = \sqrt{\frac{2 GM}{R}}$
$v_{e} = \sqrt{\frac{2 G}{R} \cdot \frac{4}{3} π R^{3} ρ} (\begin{aligned} ∵ M & = V ρ \\ = \frac{4}{3} π R^{3} ρ \end{aligned})$
$v_{e} = R \sqrt{\frac{8 G π ρ}{3}}$
$v_{e} \propto R \sqrt{ρ}$
Hence, option (b) is correct.

CG PET- 2005

Gravitation

138697 The escape velocity from the earth's surface is $11.2 Km / s$ . If a planet has a radius twice that of the earth and on which the acceleration due to gravity is twice that on the earth, then the escape velocity on this planet will be

1

11.2 km / s

2

5.6 km / s

3

22.4 km / s

4

33.6 km / s

Explanation:

C Given that,
$v_{e} = 11.2 km / s$
$R_{P} = 2 R_{e}$
$g_{P} = 2 g_{e}$
$v_{e} = \sqrt{2 gR}$
Then, $\frac{v_{P}}{v_{e}} = \frac{\sqrt{2 g_{p} R_{p}}}{\sqrt{2 g_{e} R_{e}}}$
$\frac{v_{P}}{11.2} = \frac{\sqrt{2 \times 2 g_{e} \times 2 R_{e}}}{2 g_{e} R_{e}}$
$v_{P} = 2 \times 11.2 = 22.4 km / s$

CG PET-2004

Gravitation

138698 A satellite revolves very near to the earth surface. Its speed should be around

1

5 km / s

2

8 km / s

3

2 km / s

4

11 km / s

Explanation:

B Since, orbital speed of satellite near to the earth surface is approximately $8 km / s$ .
We know that,
$orbital velocity (v) = \sqrt{\frac{GM}{R}}$
$= \sqrt{\frac{6.67 \times 10^{- 11} \times 6 \times 10^{24}}{6400 \times 1000}}$
$= 7.9 \times 10^{3} m / s$
$\approx 8 km / s$

Manipal UGET-2019

Gravitation

138699 For a given density of a planet, the orbital speed of satellite near the surface of the planet of radius $R$ is proportional to

1

R^{1 / 2}

2

R^{3 / 2}

3

R^{- 1 / 2}

4

R^{0}

Explanation:

D From the Kepler's third law -
$\frac{T^{2}}{R^{3}} = \frac{4 π^{2}}{GM}$
when $r$ is radius, semi-major axis of the elliptical near the planet surface.
$M = ρ V$
$= ρ \times \frac{4}{3} π R^{3}$
$\frac{T^{2}}{R^{3}} = \frac{4 π^{2}}{G . ρ \times \frac{4}{3} π R^{3}}$
$T = \sqrt{\frac{3 π}{G ρ}} \times R^{0}$
The planet of radius $R$ is proportional to $R^{0}$ .

Manipal UGET-2015

Gravitation

138696 If the radius of a planet is $R$ and its density is $ρ$ , the escape velocity from its surface will be

1

v_{e} \propto ρ R

2

v_{e} \propto \sqrt{ρ} R

3

v_{e} \propto \sqrt{\frac{ρ}{R}}

4

v_{e} \propto \frac{1}{\sqrt{ρ R}}

Explanation:

B We know that,
Escape velocity $(v_{e}) = \sqrt{\frac{2 GM}{R}}$
$v_{e} = \sqrt{\frac{2 G}{R} \cdot \frac{4}{3} π R^{3} ρ} (\begin{aligned} ∵ M & = V ρ \\ = \frac{4}{3} π R^{3} ρ \end{aligned})$
$v_{e} = R \sqrt{\frac{8 G π ρ}{3}}$
$v_{e} \propto R \sqrt{ρ}$
Hence, option (b) is correct.

CG PET- 2005

Gravitation

138697 The escape velocity from the earth's surface is $11.2 Km / s$ . If a planet has a radius twice that of the earth and on which the acceleration due to gravity is twice that on the earth, then the escape velocity on this planet will be

1

11.2 km / s

2

5.6 km / s

3

22.4 km / s

4

33.6 km / s

Explanation:

C Given that,
$v_{e} = 11.2 km / s$
$R_{P} = 2 R_{e}$
$g_{P} = 2 g_{e}$
$v_{e} = \sqrt{2 gR}$
Then, $\frac{v_{P}}{v_{e}} = \frac{\sqrt{2 g_{p} R_{p}}}{\sqrt{2 g_{e} R_{e}}}$
$\frac{v_{P}}{11.2} = \frac{\sqrt{2 \times 2 g_{e} \times 2 R_{e}}}{2 g_{e} R_{e}}$
$v_{P} = 2 \times 11.2 = 22.4 km / s$

CG PET-2004

Gravitation

138698 A satellite revolves very near to the earth surface. Its speed should be around

1

5 km / s

2

8 km / s

3

2 km / s

4

11 km / s

Explanation:

B Since, orbital speed of satellite near to the earth surface is approximately $8 km / s$ .
We know that,
$orbital velocity (v) = \sqrt{\frac{GM}{R}}$
$= \sqrt{\frac{6.67 \times 10^{- 11} \times 6 \times 10^{24}}{6400 \times 1000}}$
$= 7.9 \times 10^{3} m / s$
$\approx 8 km / s$

Manipal UGET-2019

Gravitation

138699 For a given density of a planet, the orbital speed of satellite near the surface of the planet of radius $R$ is proportional to

1

R^{1 / 2}

2

R^{3 / 2}

3

R^{- 1 / 2}

4

R^{0}

Explanation:

D From the Kepler's third law -
$\frac{T^{2}}{R^{3}} = \frac{4 π^{2}}{GM}$
when $r$ is radius, semi-major axis of the elliptical near the planet surface.
$M = ρ V$
$= ρ \times \frac{4}{3} π R^{3}$
$\frac{T^{2}}{R^{3}} = \frac{4 π^{2}}{G . ρ \times \frac{4}{3} π R^{3}}$
$T = \sqrt{\frac{3 π}{G ρ}} \times R^{0}$
The planet of radius $R$ is proportional to $R^{0}$ .

Manipal UGET-2015

Gravitation

138696 If the radius of a planet is $R$ and its density is $ρ$ , the escape velocity from its surface will be

1

v_{e} \propto ρ R

2

v_{e} \propto \sqrt{ρ} R

3

v_{e} \propto \sqrt{\frac{ρ}{R}}

4

v_{e} \propto \frac{1}{\sqrt{ρ R}}

Explanation:

B We know that,
Escape velocity $(v_{e}) = \sqrt{\frac{2 GM}{R}}$
$v_{e} = \sqrt{\frac{2 G}{R} \cdot \frac{4}{3} π R^{3} ρ} (\begin{aligned} ∵ M & = V ρ \\ = \frac{4}{3} π R^{3} ρ \end{aligned})$
$v_{e} = R \sqrt{\frac{8 G π ρ}{3}}$
$v_{e} \propto R \sqrt{ρ}$
Hence, option (b) is correct.

CG PET- 2005

Gravitation

138697 The escape velocity from the earth's surface is $11.2 Km / s$ . If a planet has a radius twice that of the earth and on which the acceleration due to gravity is twice that on the earth, then the escape velocity on this planet will be

1

11.2 km / s

2

5.6 km / s

3

22.4 km / s

4

33.6 km / s

Explanation:

C Given that,
$v_{e} = 11.2 km / s$
$R_{P} = 2 R_{e}$
$g_{P} = 2 g_{e}$
$v_{e} = \sqrt{2 gR}$
Then, $\frac{v_{P}}{v_{e}} = \frac{\sqrt{2 g_{p} R_{p}}}{\sqrt{2 g_{e} R_{e}}}$
$\frac{v_{P}}{11.2} = \frac{\sqrt{2 \times 2 g_{e} \times 2 R_{e}}}{2 g_{e} R_{e}}$
$v_{P} = 2 \times 11.2 = 22.4 km / s$

CG PET-2004

Gravitation

138698 A satellite revolves very near to the earth surface. Its speed should be around

1

5 km / s

2

8 km / s

3

2 km / s

4

11 km / s

Explanation:

B Since, orbital speed of satellite near to the earth surface is approximately $8 km / s$ .
We know that,
$orbital velocity (v) = \sqrt{\frac{GM}{R}}$
$= \sqrt{\frac{6.67 \times 10^{- 11} \times 6 \times 10^{24}}{6400 \times 1000}}$
$= 7.9 \times 10^{3} m / s$
$\approx 8 km / s$

Manipal UGET-2019

Gravitation

138699 For a given density of a planet, the orbital speed of satellite near the surface of the planet of radius $R$ is proportional to

1

R^{1 / 2}

2

R^{3 / 2}

3

R^{- 1 / 2}

4

R^{0}

Explanation:

D From the Kepler's third law -
$\frac{T^{2}}{R^{3}} = \frac{4 π^{2}}{GM}$
when $r$ is radius, semi-major axis of the elliptical near the planet surface.
$M = ρ V$
$= ρ \times \frac{4}{3} π R^{3}$
$\frac{T^{2}}{R^{3}} = \frac{4 π^{2}}{G . ρ \times \frac{4}{3} π R^{3}}$
$T = \sqrt{\frac{3 π}{G ρ}} \times R^{0}$
The planet of radius $R$ is proportional to $R^{0}$ .

Manipal UGET-2015