QBManager

Gravitation

138710 A mass of $6 \times 10^{24} kg$ is to be compressed in a sphere in such a way that the escape velocity from the sphere is $3 \times 10^{8} m / s$ . What should be the radius of the sphere?
$(G = 6.67 \times 10^{- 11} N - m^{2} / {kg}^{2})$

1

9 km

2

9 m

3

9 cm

4

9 mm

Explanation:

D Given that,
The mass of the sphere $= 6 \times 10^{24} kg$
$Escape velocity = 3 \times 10^{8} m / s$
$G = 6.67 \times 10^{- 11} N - m^{2} / {kg}^{2}$
We know that,
Escape velocity $(v_{e}) = \sqrt{\frac{2 GM}{R}}$
$R = \frac{2 GM}{v_{e}^{2}}$
$R = \frac{2 \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{{(3 \times 10^{8})}^{2}}$
$R = \frac{12 \times 6.67 \times 10^{13}}{9 \times 10^{16}}$
$R = \frac{80.04}{9} \times 10^{- 3}$
$R = 8.89 \times 10^{- 3} m$
$R = 9 mm$

JCECE-2010

Gravitation

138711 A satellite in a circular orbit of radius $R$ has a period of $4 h$ . Another satellite with orbital radius $3 R$ around the same planet with have a period (in hours)

1 16

2 4

3

4 \sqrt{27}

4

4 \sqrt{8}

Explanation:

C Given,
Radius of first satellite $(R_{1}) = R$
Time $(T_{1}) = 4 hr$
Radius of second satellite $(R_{2}) = 3 R$
Time $(T_{2}) =$ ?
From Kepler's law -
${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{R_{1}}{R_{2}})}^{3}$
$\frac{4}{T_{2}} = {(\frac{R}{3 R})}^{3 / 2}$
$\frac{4}{T_{2}} = {(\frac{1}{3})}^{3 / 2} \Rightarrow \frac{4}{T_{2}} = {{(\frac{1}{3})}^{3}}^{\frac{1}{2}}$
$\frac{4}{T_{2}} = \frac{1}{\sqrt{27}}$
$T_{2} = 4 \sqrt{27} hr$

JCECE-2009

Gravitation

138712 For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

1 2

2

\frac{1}{2}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

B We know that,
Kinetic energy of satellite (K.E.) $= \frac{GMm}{2 R}$
Potential energy of satellite (P.E. $) = \frac{- GMm}{R}$ Hence,
Ratio of K.E. and P.E. of satellite is -
$\frac{K . E}{P.E} = \frac{\frac{GMm}{2 R}}{\frac{GMm}{R}}$
$= \frac{GMm}{2 GMm}$
K.E : P.E $= 1 : 2$
Ratio between kinetic energy and potential energy is $1 : 2$ .

AIPMT 2005

Gravitation

138713 A body is projected vertical upwards from the surface of a planet of radius $r$ with a velocity equal to $1 / 3$ rd the escape velocity for the planet. The maximum height attained by the body is :

1

R / 2

2

R / 3

3

R / 5

4

R / 9

Explanation:

D We know that, Escape velocity $(v_{e}) = \sqrt{2 gR}$
Let $h$ be the maximum height attained these from equation of motion.
$v^{2} = u^{2} + 2 g h$
$u = 0$
$v = \sqrt{2 gh}$
From equation (i)
$∴ \sqrt{2 gh} = \frac{1}{3} \sqrt{2 gR}$
Squaring both sides of equation, we get,
$2 gh = \frac{1}{9} \times 2 gR$
$h = \frac{R}{9}$

JCECE-2005

Gravitation

138710 A mass of $6 \times 10^{24} kg$ is to be compressed in a sphere in such a way that the escape velocity from the sphere is $3 \times 10^{8} m / s$ . What should be the radius of the sphere?
$(G = 6.67 \times 10^{- 11} N - m^{2} / {kg}^{2})$

1

9 km

2

9 m

3

9 cm

4

9 mm

Explanation:

D Given that,
The mass of the sphere $= 6 \times 10^{24} kg$
$Escape velocity = 3 \times 10^{8} m / s$
$G = 6.67 \times 10^{- 11} N - m^{2} / {kg}^{2}$
We know that,
Escape velocity $(v_{e}) = \sqrt{\frac{2 GM}{R}}$
$R = \frac{2 GM}{v_{e}^{2}}$
$R = \frac{2 \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{{(3 \times 10^{8})}^{2}}$
$R = \frac{12 \times 6.67 \times 10^{13}}{9 \times 10^{16}}$
$R = \frac{80.04}{9} \times 10^{- 3}$
$R = 8.89 \times 10^{- 3} m$
$R = 9 mm$

JCECE-2010

Gravitation

138711 A satellite in a circular orbit of radius $R$ has a period of $4 h$ . Another satellite with orbital radius $3 R$ around the same planet with have a period (in hours)

1 16

2 4

3

4 \sqrt{27}

4

4 \sqrt{8}

Explanation:

C Given,
Radius of first satellite $(R_{1}) = R$
Time $(T_{1}) = 4 hr$
Radius of second satellite $(R_{2}) = 3 R$
Time $(T_{2}) =$ ?
From Kepler's law -
${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{R_{1}}{R_{2}})}^{3}$
$\frac{4}{T_{2}} = {(\frac{R}{3 R})}^{3 / 2}$
$\frac{4}{T_{2}} = {(\frac{1}{3})}^{3 / 2} \Rightarrow \frac{4}{T_{2}} = {{(\frac{1}{3})}^{3}}^{\frac{1}{2}}$
$\frac{4}{T_{2}} = \frac{1}{\sqrt{27}}$
$T_{2} = 4 \sqrt{27} hr$

JCECE-2009

Gravitation

138712 For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

1 2

2

\frac{1}{2}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

B We know that,
Kinetic energy of satellite (K.E.) $= \frac{GMm}{2 R}$
Potential energy of satellite (P.E. $) = \frac{- GMm}{R}$ Hence,
Ratio of K.E. and P.E. of satellite is -
$\frac{K . E}{P.E} = \frac{\frac{GMm}{2 R}}{\frac{GMm}{R}}$
$= \frac{GMm}{2 GMm}$
K.E : P.E $= 1 : 2$
Ratio between kinetic energy and potential energy is $1 : 2$ .

AIPMT 2005

Gravitation

138713 A body is projected vertical upwards from the surface of a planet of radius $r$ with a velocity equal to $1 / 3$ rd the escape velocity for the planet. The maximum height attained by the body is :

1

R / 2

2

R / 3

3

R / 5

4

R / 9

Explanation:

D We know that, Escape velocity $(v_{e}) = \sqrt{2 gR}$
Let $h$ be the maximum height attained these from equation of motion.
$v^{2} = u^{2} + 2 g h$
$u = 0$
$v = \sqrt{2 gh}$
From equation (i)
$∴ \sqrt{2 gh} = \frac{1}{3} \sqrt{2 gR}$
Squaring both sides of equation, we get,
$2 gh = \frac{1}{9} \times 2 gR$
$h = \frac{R}{9}$

JCECE-2005

Gravitation

138710 A mass of $6 \times 10^{24} kg$ is to be compressed in a sphere in such a way that the escape velocity from the sphere is $3 \times 10^{8} m / s$ . What should be the radius of the sphere?
$(G = 6.67 \times 10^{- 11} N - m^{2} / {kg}^{2})$

1

9 km

2

9 m

3

9 cm

4

9 mm

Explanation:

D Given that,
The mass of the sphere $= 6 \times 10^{24} kg$
$Escape velocity = 3 \times 10^{8} m / s$
$G = 6.67 \times 10^{- 11} N - m^{2} / {kg}^{2}$
We know that,
Escape velocity $(v_{e}) = \sqrt{\frac{2 GM}{R}}$
$R = \frac{2 GM}{v_{e}^{2}}$
$R = \frac{2 \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{{(3 \times 10^{8})}^{2}}$
$R = \frac{12 \times 6.67 \times 10^{13}}{9 \times 10^{16}}$
$R = \frac{80.04}{9} \times 10^{- 3}$
$R = 8.89 \times 10^{- 3} m$
$R = 9 mm$

JCECE-2010

Gravitation

138711 A satellite in a circular orbit of radius $R$ has a period of $4 h$ . Another satellite with orbital radius $3 R$ around the same planet with have a period (in hours)

1 16

2 4

3

4 \sqrt{27}

4

4 \sqrt{8}

Explanation:

C Given,
Radius of first satellite $(R_{1}) = R$
Time $(T_{1}) = 4 hr$
Radius of second satellite $(R_{2}) = 3 R$
Time $(T_{2}) =$ ?
From Kepler's law -
${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{R_{1}}{R_{2}})}^{3}$
$\frac{4}{T_{2}} = {(\frac{R}{3 R})}^{3 / 2}$
$\frac{4}{T_{2}} = {(\frac{1}{3})}^{3 / 2} \Rightarrow \frac{4}{T_{2}} = {{(\frac{1}{3})}^{3}}^{\frac{1}{2}}$
$\frac{4}{T_{2}} = \frac{1}{\sqrt{27}}$
$T_{2} = 4 \sqrt{27} hr$

JCECE-2009

Gravitation

138712 For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

1 2

2

\frac{1}{2}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

B We know that,
Kinetic energy of satellite (K.E.) $= \frac{GMm}{2 R}$
Potential energy of satellite (P.E. $) = \frac{- GMm}{R}$ Hence,
Ratio of K.E. and P.E. of satellite is -
$\frac{K . E}{P.E} = \frac{\frac{GMm}{2 R}}{\frac{GMm}{R}}$
$= \frac{GMm}{2 GMm}$
K.E : P.E $= 1 : 2$
Ratio between kinetic energy and potential energy is $1 : 2$ .

AIPMT 2005

Gravitation

138713 A body is projected vertical upwards from the surface of a planet of radius $r$ with a velocity equal to $1 / 3$ rd the escape velocity for the planet. The maximum height attained by the body is :

1

R / 2

2

R / 3

3

R / 5

4

R / 9

Explanation:

D We know that, Escape velocity $(v_{e}) = \sqrt{2 gR}$
Let $h$ be the maximum height attained these from equation of motion.
$v^{2} = u^{2} + 2 g h$
$u = 0$
$v = \sqrt{2 gh}$
From equation (i)
$∴ \sqrt{2 gh} = \frac{1}{3} \sqrt{2 gR}$
Squaring both sides of equation, we get,
$2 gh = \frac{1}{9} \times 2 gR$
$h = \frac{R}{9}$

JCECE-2005

Gravitation

138710 A mass of $6 \times 10^{24} kg$ is to be compressed in a sphere in such a way that the escape velocity from the sphere is $3 \times 10^{8} m / s$ . What should be the radius of the sphere?
$(G = 6.67 \times 10^{- 11} N - m^{2} / {kg}^{2})$

1

9 km

2

9 m

3

9 cm

4

9 mm

Explanation:

D Given that,
The mass of the sphere $= 6 \times 10^{24} kg$
$Escape velocity = 3 \times 10^{8} m / s$
$G = 6.67 \times 10^{- 11} N - m^{2} / {kg}^{2}$
We know that,
Escape velocity $(v_{e}) = \sqrt{\frac{2 GM}{R}}$
$R = \frac{2 GM}{v_{e}^{2}}$
$R = \frac{2 \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{{(3 \times 10^{8})}^{2}}$
$R = \frac{12 \times 6.67 \times 10^{13}}{9 \times 10^{16}}$
$R = \frac{80.04}{9} \times 10^{- 3}$
$R = 8.89 \times 10^{- 3} m$
$R = 9 mm$

JCECE-2010

Gravitation

138711 A satellite in a circular orbit of radius $R$ has a period of $4 h$ . Another satellite with orbital radius $3 R$ around the same planet with have a period (in hours)

1 16

2 4

3

4 \sqrt{27}

4

4 \sqrt{8}

Explanation:

C Given,
Radius of first satellite $(R_{1}) = R$
Time $(T_{1}) = 4 hr$
Radius of second satellite $(R_{2}) = 3 R$
Time $(T_{2}) =$ ?
From Kepler's law -
${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{R_{1}}{R_{2}})}^{3}$
$\frac{4}{T_{2}} = {(\frac{R}{3 R})}^{3 / 2}$
$\frac{4}{T_{2}} = {(\frac{1}{3})}^{3 / 2} \Rightarrow \frac{4}{T_{2}} = {{(\frac{1}{3})}^{3}}^{\frac{1}{2}}$
$\frac{4}{T_{2}} = \frac{1}{\sqrt{27}}$
$T_{2} = 4 \sqrt{27} hr$

JCECE-2009

Gravitation

138712 For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

1 2

2

\frac{1}{2}

3

\frac{1}{\sqrt{2}}

4

\sqrt{2}

Explanation:

B We know that,
Kinetic energy of satellite (K.E.) $= \frac{GMm}{2 R}$
Potential energy of satellite (P.E. $) = \frac{- GMm}{R}$ Hence,
Ratio of K.E. and P.E. of satellite is -
$\frac{K . E}{P.E} = \frac{\frac{GMm}{2 R}}{\frac{GMm}{R}}$
$= \frac{GMm}{2 GMm}$
K.E : P.E $= 1 : 2$
Ratio between kinetic energy and potential energy is $1 : 2$ .

AIPMT 2005

Gravitation

138713 A body is projected vertical upwards from the surface of a planet of radius $r$ with a velocity equal to $1 / 3$ rd the escape velocity for the planet. The maximum height attained by the body is :

1

R / 2

2

R / 3

3

R / 5

4

R / 9

Explanation:

D We know that, Escape velocity $(v_{e}) = \sqrt{2 gR}$
Let $h$ be the maximum height attained these from equation of motion.
$v^{2} = u^{2} + 2 g h$
$u = 0$
$v = \sqrt{2 gh}$
From equation (i)
$∴ \sqrt{2 gh} = \frac{1}{3} \sqrt{2 gR}$
Squaring both sides of equation, we get,
$2 gh = \frac{1}{9} \times 2 gR$
$h = \frac{R}{9}$

JCECE-2005

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