QBManager

Gravitation

138676 A satellite is launched in a circular orbit of radius $R$ around the earth. $A$ second satellite is launched into an orbit or radius $1.01 R$ . The time period of second satellite is longer than the first one (approximately) by

1

1.5 %

2

0.5 %

3

3 %

4

1 %

5

2 %

Explanation:

A Given,
$R_{1} = R, R_{2} = 1.01 R$
We know that,
From Kepler's law,
$T^{2} \propto R^{3}$
${(\frac{T_{2}}{T_{1}})}^{2} = {(\frac{R_{2}}{R_{1}})}^{3}$
${(\frac{T_{2}}{T_{1}})}^{2} = {(\frac{1.01 R}{R})}^{3} = (1 + 0.01)^{3}$
By Binomial Expression
$\frac{T_{2}}{T_{1}} = (1 + 0.01)^{3 / 2} = (1 + \frac{3}{2} \times 0.001)$
$\frac{T_{2}}{T_{1}} = 1.015$
$\frac{T_{2} - T_{1}}{T_{1}} \times 100 = (1.015 - 1) \times 100$
$= 1.5 %$

AIIMS 2002

Gravitation

138677 The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $v$ , then the escape velocity from the planet is:

1

\sqrt{3} v

2

\sqrt{2} v

3

V

4

\sqrt{5} v

5

\sqrt{12} v

Explanation:

A Let $M_{p}$ and $M_{e}$ are the mass of planet and earth and $R_{P}$ and $R_{e}$ are the radius of planet and earth respectively.
Then $M_{p} = 6 M_{e}, R_{p} = 2 R_{e}$
We know that, escape velocity of earth -
$∵ v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}$
Escape velocity from the planet,
$v_{P} = \sqrt{\frac{2 {GM}_{P}}{R_{P}}}$
$v_{P} = \sqrt{\frac{2 G \times 6 M_{e}}{2 R_{e}}}$
$V_{P} = \sqrt{\frac{6 {GM}_{e}}{R_{e}}}$
$= \sqrt{\frac{2 \times 3 \times {GM}_{e}}{R_{e}}}$
$v_{P} = \sqrt{3} v_{e}$
${∵ v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}}$
$v_{P} = \sqrt{3} v$

Kerala CEE 2006

Gravitation

138678 The escape velocity for the earth is $v_{e^{*}}$ . The escape velocity for a planet whose radius is $\frac{1}{4}$ th the radius of the earth and mass half that of the earth is

1

\frac{V_{e}}{\sqrt{2}}

2

\sqrt{2} v_{e}

3

2 v_{e}

4

\frac{v_{e}}{2}

Explanation:

B Given,
$Radius of planet (R_{p}) = \frac{R_{e}}{4}$
$Mass of planet (M_{p}) = \frac{M_{e}}{2}$
Escape velocity of a body from the surface of the earth is given by
$v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}$
Then escape velocity of a body from the surface of the planet is given by
$v_{p} = \sqrt{\frac{2 {GM}_{p}}{R_{p}}}$
$= \sqrt{\frac{2 G \times M_{e} \times 4}{2 R_{e}}}$
$= \sqrt{\frac{2 {GM}_{e}}{R_{e}}} \times \sqrt{2}$
From equation (i) we get -
$= \sqrt{2} v_{e}$

UPSEE - 2014

Gravitation

138679 The velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be

1

\sqrt{gR}

2

\sqrt{0.5 gR}

3

\sqrt{2 gR}

4

\sqrt{3 g R}

Explanation:

A According to law of universal gravity
$F = \frac{GMm}{R^{2}}$
And,
$F = mg$ (According to Newton's second law of motion
$\frac{GMm}{R^{2}} = mg$
$\frac{GM}{R^{2}} = g$
We know that orbital velocity
$v = \sqrt{\frac{GM}{R}}$
$v = \sqrt{\frac{{gR}^{2}}{R}} = \sqrt{gR}$
$v = \sqrt{gR}$
Hence, the velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be $\sqrt{gR}$ .

UPSEE - 2011

Gravitation

138676 A satellite is launched in a circular orbit of radius $R$ around the earth. $A$ second satellite is launched into an orbit or radius $1.01 R$ . The time period of second satellite is longer than the first one (approximately) by

1

1.5 %

2

0.5 %

3

3 %

4

1 %

5

2 %

Explanation:

A Given,
$R_{1} = R, R_{2} = 1.01 R$
We know that,
From Kepler's law,
$T^{2} \propto R^{3}$
${(\frac{T_{2}}{T_{1}})}^{2} = {(\frac{R_{2}}{R_{1}})}^{3}$
${(\frac{T_{2}}{T_{1}})}^{2} = {(\frac{1.01 R}{R})}^{3} = (1 + 0.01)^{3}$
By Binomial Expression
$\frac{T_{2}}{T_{1}} = (1 + 0.01)^{3 / 2} = (1 + \frac{3}{2} \times 0.001)$
$\frac{T_{2}}{T_{1}} = 1.015$
$\frac{T_{2} - T_{1}}{T_{1}} \times 100 = (1.015 - 1) \times 100$
$= 1.5 %$

AIIMS 2002

Gravitation

138677 The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $v$ , then the escape velocity from the planet is:

1

\sqrt{3} v

2

\sqrt{2} v

3

V

4

\sqrt{5} v

5

\sqrt{12} v

Explanation:

A Let $M_{p}$ and $M_{e}$ are the mass of planet and earth and $R_{P}$ and $R_{e}$ are the radius of planet and earth respectively.
Then $M_{p} = 6 M_{e}, R_{p} = 2 R_{e}$
We know that, escape velocity of earth -
$∵ v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}$
Escape velocity from the planet,
$v_{P} = \sqrt{\frac{2 {GM}_{P}}{R_{P}}}$
$v_{P} = \sqrt{\frac{2 G \times 6 M_{e}}{2 R_{e}}}$
$V_{P} = \sqrt{\frac{6 {GM}_{e}}{R_{e}}}$
$= \sqrt{\frac{2 \times 3 \times {GM}_{e}}{R_{e}}}$
$v_{P} = \sqrt{3} v_{e}$
${∵ v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}}$
$v_{P} = \sqrt{3} v$

Kerala CEE 2006

Gravitation

138678 The escape velocity for the earth is $v_{e^{*}}$ . The escape velocity for a planet whose radius is $\frac{1}{4}$ th the radius of the earth and mass half that of the earth is

1

\frac{V_{e}}{\sqrt{2}}

2

\sqrt{2} v_{e}

3

2 v_{e}

4

\frac{v_{e}}{2}

Explanation:

B Given,
$Radius of planet (R_{p}) = \frac{R_{e}}{4}$
$Mass of planet (M_{p}) = \frac{M_{e}}{2}$
Escape velocity of a body from the surface of the earth is given by
$v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}$
Then escape velocity of a body from the surface of the planet is given by
$v_{p} = \sqrt{\frac{2 {GM}_{p}}{R_{p}}}$
$= \sqrt{\frac{2 G \times M_{e} \times 4}{2 R_{e}}}$
$= \sqrt{\frac{2 {GM}_{e}}{R_{e}}} \times \sqrt{2}$
From equation (i) we get -
$= \sqrt{2} v_{e}$

UPSEE - 2014

Gravitation

138679 The velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be

1

\sqrt{gR}

2

\sqrt{0.5 gR}

3

\sqrt{2 gR}

4

\sqrt{3 g R}

Explanation:

A According to law of universal gravity
$F = \frac{GMm}{R^{2}}$
And,
$F = mg$ (According to Newton's second law of motion
$\frac{GMm}{R^{2}} = mg$
$\frac{GM}{R^{2}} = g$
We know that orbital velocity
$v = \sqrt{\frac{GM}{R}}$
$v = \sqrt{\frac{{gR}^{2}}{R}} = \sqrt{gR}$
$v = \sqrt{gR}$
Hence, the velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be $\sqrt{gR}$ .

UPSEE - 2011

Gravitation

138676 A satellite is launched in a circular orbit of radius $R$ around the earth. $A$ second satellite is launched into an orbit or radius $1.01 R$ . The time period of second satellite is longer than the first one (approximately) by

1

1.5 %

2

0.5 %

3

3 %

4

1 %

5

2 %

Explanation:

A Given,
$R_{1} = R, R_{2} = 1.01 R$
We know that,
From Kepler's law,
$T^{2} \propto R^{3}$
${(\frac{T_{2}}{T_{1}})}^{2} = {(\frac{R_{2}}{R_{1}})}^{3}$
${(\frac{T_{2}}{T_{1}})}^{2} = {(\frac{1.01 R}{R})}^{3} = (1 + 0.01)^{3}$
By Binomial Expression
$\frac{T_{2}}{T_{1}} = (1 + 0.01)^{3 / 2} = (1 + \frac{3}{2} \times 0.001)$
$\frac{T_{2}}{T_{1}} = 1.015$
$\frac{T_{2} - T_{1}}{T_{1}} \times 100 = (1.015 - 1) \times 100$
$= 1.5 %$

AIIMS 2002

Gravitation

138677 The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $v$ , then the escape velocity from the planet is:

1

\sqrt{3} v

2

\sqrt{2} v

3

V

4

\sqrt{5} v

5

\sqrt{12} v

Explanation:

A Let $M_{p}$ and $M_{e}$ are the mass of planet and earth and $R_{P}$ and $R_{e}$ are the radius of planet and earth respectively.
Then $M_{p} = 6 M_{e}, R_{p} = 2 R_{e}$
We know that, escape velocity of earth -
$∵ v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}$
Escape velocity from the planet,
$v_{P} = \sqrt{\frac{2 {GM}_{P}}{R_{P}}}$
$v_{P} = \sqrt{\frac{2 G \times 6 M_{e}}{2 R_{e}}}$
$V_{P} = \sqrt{\frac{6 {GM}_{e}}{R_{e}}}$
$= \sqrt{\frac{2 \times 3 \times {GM}_{e}}{R_{e}}}$
$v_{P} = \sqrt{3} v_{e}$
${∵ v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}}$
$v_{P} = \sqrt{3} v$

Kerala CEE 2006

Gravitation

138678 The escape velocity for the earth is $v_{e^{*}}$ . The escape velocity for a planet whose radius is $\frac{1}{4}$ th the radius of the earth and mass half that of the earth is

1

\frac{V_{e}}{\sqrt{2}}

2

\sqrt{2} v_{e}

3

2 v_{e}

4

\frac{v_{e}}{2}

Explanation:

B Given,
$Radius of planet (R_{p}) = \frac{R_{e}}{4}$
$Mass of planet (M_{p}) = \frac{M_{e}}{2}$
Escape velocity of a body from the surface of the earth is given by
$v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}$
Then escape velocity of a body from the surface of the planet is given by
$v_{p} = \sqrt{\frac{2 {GM}_{p}}{R_{p}}}$
$= \sqrt{\frac{2 G \times M_{e} \times 4}{2 R_{e}}}$
$= \sqrt{\frac{2 {GM}_{e}}{R_{e}}} \times \sqrt{2}$
From equation (i) we get -
$= \sqrt{2} v_{e}$

UPSEE - 2014

Gravitation

138679 The velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be

1

\sqrt{gR}

2

\sqrt{0.5 gR}

3

\sqrt{2 gR}

4

\sqrt{3 g R}

Explanation:

A According to law of universal gravity
$F = \frac{GMm}{R^{2}}$
And,
$F = mg$ (According to Newton's second law of motion
$\frac{GMm}{R^{2}} = mg$
$\frac{GM}{R^{2}} = g$
We know that orbital velocity
$v = \sqrt{\frac{GM}{R}}$
$v = \sqrt{\frac{{gR}^{2}}{R}} = \sqrt{gR}$
$v = \sqrt{gR}$
Hence, the velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be $\sqrt{gR}$ .

UPSEE - 2011

Gravitation

138676 A satellite is launched in a circular orbit of radius $R$ around the earth. $A$ second satellite is launched into an orbit or radius $1.01 R$ . The time period of second satellite is longer than the first one (approximately) by

1

1.5 %

2

0.5 %

3

3 %

4

1 %

5

2 %

Explanation:

A Given,
$R_{1} = R, R_{2} = 1.01 R$
We know that,
From Kepler's law,
$T^{2} \propto R^{3}$
${(\frac{T_{2}}{T_{1}})}^{2} = {(\frac{R_{2}}{R_{1}})}^{3}$
${(\frac{T_{2}}{T_{1}})}^{2} = {(\frac{1.01 R}{R})}^{3} = (1 + 0.01)^{3}$
By Binomial Expression
$\frac{T_{2}}{T_{1}} = (1 + 0.01)^{3 / 2} = (1 + \frac{3}{2} \times 0.001)$
$\frac{T_{2}}{T_{1}} = 1.015$
$\frac{T_{2} - T_{1}}{T_{1}} \times 100 = (1.015 - 1) \times 100$
$= 1.5 %$

AIIMS 2002

Gravitation

138677 The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is $v$ , then the escape velocity from the planet is:

1

\sqrt{3} v

2

\sqrt{2} v

3

V

4

\sqrt{5} v

5

\sqrt{12} v

Explanation:

A Let $M_{p}$ and $M_{e}$ are the mass of planet and earth and $R_{P}$ and $R_{e}$ are the radius of planet and earth respectively.
Then $M_{p} = 6 M_{e}, R_{p} = 2 R_{e}$
We know that, escape velocity of earth -
$∵ v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}$
Escape velocity from the planet,
$v_{P} = \sqrt{\frac{2 {GM}_{P}}{R_{P}}}$
$v_{P} = \sqrt{\frac{2 G \times 6 M_{e}}{2 R_{e}}}$
$V_{P} = \sqrt{\frac{6 {GM}_{e}}{R_{e}}}$
$= \sqrt{\frac{2 \times 3 \times {GM}_{e}}{R_{e}}}$
$v_{P} = \sqrt{3} v_{e}$
${∵ v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}}$
$v_{P} = \sqrt{3} v$

Kerala CEE 2006

Gravitation

138678 The escape velocity for the earth is $v_{e^{*}}$ . The escape velocity for a planet whose radius is $\frac{1}{4}$ th the radius of the earth and mass half that of the earth is

1

\frac{V_{e}}{\sqrt{2}}

2

\sqrt{2} v_{e}

3

2 v_{e}

4

\frac{v_{e}}{2}

Explanation:

B Given,
$Radius of planet (R_{p}) = \frac{R_{e}}{4}$
$Mass of planet (M_{p}) = \frac{M_{e}}{2}$
Escape velocity of a body from the surface of the earth is given by
$v_{e} = \sqrt{\frac{2 {GM}_{e}}{R_{e}}}$
Then escape velocity of a body from the surface of the planet is given by
$v_{p} = \sqrt{\frac{2 {GM}_{p}}{R_{p}}}$
$= \sqrt{\frac{2 G \times M_{e} \times 4}{2 R_{e}}}$
$= \sqrt{\frac{2 {GM}_{e}}{R_{e}}} \times \sqrt{2}$
From equation (i) we get -
$= \sqrt{2} v_{e}$

UPSEE - 2014

Gravitation

138679 The velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be

1

\sqrt{gR}

2

\sqrt{0.5 gR}

3

\sqrt{2 gR}

4

\sqrt{3 g R}

Explanation:

A According to law of universal gravity
$F = \frac{GMm}{R^{2}}$
And,
$F = mg$ (According to Newton's second law of motion
$\frac{GMm}{R^{2}} = mg$
$\frac{GM}{R^{2}} = g$
We know that orbital velocity
$v = \sqrt{\frac{GM}{R}}$
$v = \sqrt{\frac{{gR}^{2}}{R}} = \sqrt{gR}$
$v = \sqrt{gR}$
Hence, the velocity of a satellite moving in an orbit about earth at a distance equal to radius $R$ of earth will be $\sqrt{gR}$ .

UPSEE - 2011

Question Bank Series

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Question Bank Series

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