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PHXI08:GRAVITATION

359813 The mass of the moon is $\frac{1}{144}$ times the mass ofa planet and its diameter is $\frac{1}{16}$ times the diameter of a planet. If the escape velocity on the planet is $v$ , the escape velocity on the moon will be

1

\frac{v}{4}

2

\frac{v}{12}

3

\frac{v}{3}

4

\frac{v}{6}

Explanation:

Escapee velocity on the planet,
$v = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{4 G M}{D}} (1)$
Where, $M =$ Mass of the planet
$R =$ Radius of the planet
$G =$ Universal gravitational constant
$D =$ Diameter of the planet
$v \propto \sqrt{\frac{M}{D}}$
For moon, $v^{'} = \sqrt{\frac{2 G M^{'}}{R^{'}}} = \sqrt{\frac{4 G M^{'}}{D^{'}}} (2)$
where, $M^{'} =$ mass of the moon
$R^{'} =$ Radius of the moon;
$D^{'} =$ diameter of the moon
Given : $M^{'} = \frac{M}{144}$ and $D^{'} = \frac{D}{16}$
From equation (1) and (2), we get
$\frac{v^{'}}{v} = \sqrt{\frac{M^{'} D}{D^{'} M}}; \frac{v^{'}}{v} = \sqrt{\frac{\frac{M}{\frac{144}{D}} \times \frac{D}{1}}{16} \times \frac{M}{1}}$
$= \frac{1}{3} \Rightarrow v^{'} = \frac{v}{3}$

JEE - 2024

PHXI08:GRAVITATION

359814 A stationary object is released from a point $P$ at a distance $3 R$ from the centre of the Moon which has radius $R$ and Mass $M$ . Which one of the following expressions gives the speed of the object on hitting the Moon?
supporting img

1

{(\frac{2 G M}{3 R})}^{1 / 2}

2

{(\frac{4 G M}{3 R})}^{1 / 2}

3

{(\frac{2 G M}{R})}^{1 / 2}

4

{(\frac{G M}{R})}^{1 / 2}

Explanation:

$U_{i} + K_{i} = U_{f} + K_{f}$
$\begin{aligned} \frac{- G M m}{3 R} + 0 = \frac{- G M m}{R} + \frac{1}{2} m v_{f}^{2} \\ \Rightarrow \frac{1}{2} m v_{f}^{2} = \frac{G M m}{R} - \frac{G M m}{3 R} \end{aligned}$
$\frac{1}{2} v_{f}^{2} = \frac{2}{3} \frac{G M}{R} \Rightarrow v_{f} = {(\frac{4}{3} \frac{G M}{R})}^{\frac{1}{2}}$

PHXI08:GRAVITATION

359815 The escape velocity from the surface of the Earth is (where $R_{E}$ is the radius of the earth)

1

\sqrt{2 g R_{E}}

2

\sqrt{g R_{E}}

3

2 \sqrt{g R_{E}}

4

\sqrt{3 g R_{E}}

Explanation:

The escape velocity form the surface of the Earth is
$v_{e} = \sqrt{\frac{2 G M_{E}}{R_{E}}} = \sqrt{2 g R_{E}} (∵ g = \frac{G M_{E}}{R_{E}^{2}})$

PHXI08:GRAVITATION

359816 There is no atmosphere on moon because

1 It revolves round the earth

2 It is closer to earth

3 The escape velocity of the gas molecules is less than their root mean square velocity here

4 None of the above

Explanation:

As, the escape velocity on the surface of moon's is small i.e., 2.4 $k m / s$ in comparision to 11.2 $k m / s$ that of earth and also this velocity is smaller than $r m s$ velocity of gas molecules, therefore gas molecules escape moon's gravitation. Thus, moon cannot have atmosphere.

PHXI08:GRAVITATION

359817 A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energies is

1 Positive

2 Negative

3 Zero

4 May be positive or negative depending upon its initial velocity

Explanation:

When a body is projected with escape velocity then $T . E$ is zero at infinity.
$\frac{- G M m}{R} + \frac{1}{2} m v_{e}^{2} = 0$
If the initial velocity is less than $v_{e}$ then the body is in bound with earth and hence $T . E$ $= (-)$ ve

PHXI08:GRAVITATION

359813 The mass of the moon is $\frac{1}{144}$ times the mass ofa planet and its diameter is $\frac{1}{16}$ times the diameter of a planet. If the escape velocity on the planet is $v$ , the escape velocity on the moon will be

1

\frac{v}{4}

2

\frac{v}{12}

3

\frac{v}{3}

4

\frac{v}{6}

Explanation:

Escapee velocity on the planet,
$v = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{4 G M}{D}} (1)$
Where, $M =$ Mass of the planet
$R =$ Radius of the planet
$G =$ Universal gravitational constant
$D =$ Diameter of the planet
$v \propto \sqrt{\frac{M}{D}}$
For moon, $v^{'} = \sqrt{\frac{2 G M^{'}}{R^{'}}} = \sqrt{\frac{4 G M^{'}}{D^{'}}} (2)$
where, $M^{'} =$ mass of the moon
$R^{'} =$ Radius of the moon;
$D^{'} =$ diameter of the moon
Given : $M^{'} = \frac{M}{144}$ and $D^{'} = \frac{D}{16}$
From equation (1) and (2), we get
$\frac{v^{'}}{v} = \sqrt{\frac{M^{'} D}{D^{'} M}}; \frac{v^{'}}{v} = \sqrt{\frac{\frac{M}{\frac{144}{D}} \times \frac{D}{1}}{16} \times \frac{M}{1}}$
$= \frac{1}{3} \Rightarrow v^{'} = \frac{v}{3}$

JEE - 2024

PHXI08:GRAVITATION

359814 A stationary object is released from a point $P$ at a distance $3 R$ from the centre of the Moon which has radius $R$ and Mass $M$ . Which one of the following expressions gives the speed of the object on hitting the Moon?
supporting img

1

{(\frac{2 G M}{3 R})}^{1 / 2}

2

{(\frac{4 G M}{3 R})}^{1 / 2}

3

{(\frac{2 G M}{R})}^{1 / 2}

4

{(\frac{G M}{R})}^{1 / 2}

Explanation:

$U_{i} + K_{i} = U_{f} + K_{f}$
$\begin{aligned} \frac{- G M m}{3 R} + 0 = \frac{- G M m}{R} + \frac{1}{2} m v_{f}^{2} \\ \Rightarrow \frac{1}{2} m v_{f}^{2} = \frac{G M m}{R} - \frac{G M m}{3 R} \end{aligned}$
$\frac{1}{2} v_{f}^{2} = \frac{2}{3} \frac{G M}{R} \Rightarrow v_{f} = {(\frac{4}{3} \frac{G M}{R})}^{\frac{1}{2}}$

PHXI08:GRAVITATION

359815 The escape velocity from the surface of the Earth is (where $R_{E}$ is the radius of the earth)

1

\sqrt{2 g R_{E}}

2

\sqrt{g R_{E}}

3

2 \sqrt{g R_{E}}

4

\sqrt{3 g R_{E}}

Explanation:

The escape velocity form the surface of the Earth is
$v_{e} = \sqrt{\frac{2 G M_{E}}{R_{E}}} = \sqrt{2 g R_{E}} (∵ g = \frac{G M_{E}}{R_{E}^{2}})$

PHXI08:GRAVITATION

359816 There is no atmosphere on moon because

1 It revolves round the earth

2 It is closer to earth

3 The escape velocity of the gas molecules is less than their root mean square velocity here

4 None of the above

Explanation:

As, the escape velocity on the surface of moon's is small i.e., 2.4 $k m / s$ in comparision to 11.2 $k m / s$ that of earth and also this velocity is smaller than $r m s$ velocity of gas molecules, therefore gas molecules escape moon's gravitation. Thus, moon cannot have atmosphere.

PHXI08:GRAVITATION

359817 A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energies is

1 Positive

2 Negative

3 Zero

4 May be positive or negative depending upon its initial velocity

Explanation:

When a body is projected with escape velocity then $T . E$ is zero at infinity.
$\frac{- G M m}{R} + \frac{1}{2} m v_{e}^{2} = 0$
If the initial velocity is less than $v_{e}$ then the body is in bound with earth and hence $T . E$ $= (-)$ ve

PHXI08:GRAVITATION

359813 The mass of the moon is $\frac{1}{144}$ times the mass ofa planet and its diameter is $\frac{1}{16}$ times the diameter of a planet. If the escape velocity on the planet is $v$ , the escape velocity on the moon will be

1

\frac{v}{4}

2

\frac{v}{12}

3

\frac{v}{3}

4

\frac{v}{6}

Explanation:

Escapee velocity on the planet,
$v = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{4 G M}{D}} (1)$
Where, $M =$ Mass of the planet
$R =$ Radius of the planet
$G =$ Universal gravitational constant
$D =$ Diameter of the planet
$v \propto \sqrt{\frac{M}{D}}$
For moon, $v^{'} = \sqrt{\frac{2 G M^{'}}{R^{'}}} = \sqrt{\frac{4 G M^{'}}{D^{'}}} (2)$
where, $M^{'} =$ mass of the moon
$R^{'} =$ Radius of the moon;
$D^{'} =$ diameter of the moon
Given : $M^{'} = \frac{M}{144}$ and $D^{'} = \frac{D}{16}$
From equation (1) and (2), we get
$\frac{v^{'}}{v} = \sqrt{\frac{M^{'} D}{D^{'} M}}; \frac{v^{'}}{v} = \sqrt{\frac{\frac{M}{\frac{144}{D}} \times \frac{D}{1}}{16} \times \frac{M}{1}}$
$= \frac{1}{3} \Rightarrow v^{'} = \frac{v}{3}$

JEE - 2024

PHXI08:GRAVITATION

359814 A stationary object is released from a point $P$ at a distance $3 R$ from the centre of the Moon which has radius $R$ and Mass $M$ . Which one of the following expressions gives the speed of the object on hitting the Moon?
supporting img

1

{(\frac{2 G M}{3 R})}^{1 / 2}

2

{(\frac{4 G M}{3 R})}^{1 / 2}

3

{(\frac{2 G M}{R})}^{1 / 2}

4

{(\frac{G M}{R})}^{1 / 2}

Explanation:

$U_{i} + K_{i} = U_{f} + K_{f}$
$\begin{aligned} \frac{- G M m}{3 R} + 0 = \frac{- G M m}{R} + \frac{1}{2} m v_{f}^{2} \\ \Rightarrow \frac{1}{2} m v_{f}^{2} = \frac{G M m}{R} - \frac{G M m}{3 R} \end{aligned}$
$\frac{1}{2} v_{f}^{2} = \frac{2}{3} \frac{G M}{R} \Rightarrow v_{f} = {(\frac{4}{3} \frac{G M}{R})}^{\frac{1}{2}}$

PHXI08:GRAVITATION

359815 The escape velocity from the surface of the Earth is (where $R_{E}$ is the radius of the earth)

1

\sqrt{2 g R_{E}}

2

\sqrt{g R_{E}}

3

2 \sqrt{g R_{E}}

4

\sqrt{3 g R_{E}}

Explanation:

The escape velocity form the surface of the Earth is
$v_{e} = \sqrt{\frac{2 G M_{E}}{R_{E}}} = \sqrt{2 g R_{E}} (∵ g = \frac{G M_{E}}{R_{E}^{2}})$

PHXI08:GRAVITATION

359816 There is no atmosphere on moon because

1 It revolves round the earth

2 It is closer to earth

3 The escape velocity of the gas molecules is less than their root mean square velocity here

4 None of the above

Explanation:

As, the escape velocity on the surface of moon's is small i.e., 2.4 $k m / s$ in comparision to 11.2 $k m / s$ that of earth and also this velocity is smaller than $r m s$ velocity of gas molecules, therefore gas molecules escape moon's gravitation. Thus, moon cannot have atmosphere.

PHXI08:GRAVITATION

359817 A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energies is

1 Positive

2 Negative

3 Zero

4 May be positive or negative depending upon its initial velocity

Explanation:

When a body is projected with escape velocity then $T . E$ is zero at infinity.
$\frac{- G M m}{R} + \frac{1}{2} m v_{e}^{2} = 0$
If the initial velocity is less than $v_{e}$ then the body is in bound with earth and hence $T . E$ $= (-)$ ve

PHXI08:GRAVITATION

359813 The mass of the moon is $\frac{1}{144}$ times the mass ofa planet and its diameter is $\frac{1}{16}$ times the diameter of a planet. If the escape velocity on the planet is $v$ , the escape velocity on the moon will be

1

\frac{v}{4}

2

\frac{v}{12}

3

\frac{v}{3}

4

\frac{v}{6}

Explanation:

Escapee velocity on the planet,
$v = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{4 G M}{D}} (1)$
Where, $M =$ Mass of the planet
$R =$ Radius of the planet
$G =$ Universal gravitational constant
$D =$ Diameter of the planet
$v \propto \sqrt{\frac{M}{D}}$
For moon, $v^{'} = \sqrt{\frac{2 G M^{'}}{R^{'}}} = \sqrt{\frac{4 G M^{'}}{D^{'}}} (2)$
where, $M^{'} =$ mass of the moon
$R^{'} =$ Radius of the moon;
$D^{'} =$ diameter of the moon
Given : $M^{'} = \frac{M}{144}$ and $D^{'} = \frac{D}{16}$
From equation (1) and (2), we get
$\frac{v^{'}}{v} = \sqrt{\frac{M^{'} D}{D^{'} M}}; \frac{v^{'}}{v} = \sqrt{\frac{\frac{M}{\frac{144}{D}} \times \frac{D}{1}}{16} \times \frac{M}{1}}$
$= \frac{1}{3} \Rightarrow v^{'} = \frac{v}{3}$

JEE - 2024

PHXI08:GRAVITATION

359814 A stationary object is released from a point $P$ at a distance $3 R$ from the centre of the Moon which has radius $R$ and Mass $M$ . Which one of the following expressions gives the speed of the object on hitting the Moon?
supporting img

1

{(\frac{2 G M}{3 R})}^{1 / 2}

2

{(\frac{4 G M}{3 R})}^{1 / 2}

3

{(\frac{2 G M}{R})}^{1 / 2}

4

{(\frac{G M}{R})}^{1 / 2}

Explanation:

$U_{i} + K_{i} = U_{f} + K_{f}$
$\begin{aligned} \frac{- G M m}{3 R} + 0 = \frac{- G M m}{R} + \frac{1}{2} m v_{f}^{2} \\ \Rightarrow \frac{1}{2} m v_{f}^{2} = \frac{G M m}{R} - \frac{G M m}{3 R} \end{aligned}$
$\frac{1}{2} v_{f}^{2} = \frac{2}{3} \frac{G M}{R} \Rightarrow v_{f} = {(\frac{4}{3} \frac{G M}{R})}^{\frac{1}{2}}$

PHXI08:GRAVITATION

359815 The escape velocity from the surface of the Earth is (where $R_{E}$ is the radius of the earth)

1

\sqrt{2 g R_{E}}

2

\sqrt{g R_{E}}

3

2 \sqrt{g R_{E}}

4

\sqrt{3 g R_{E}}

Explanation:

The escape velocity form the surface of the Earth is
$v_{e} = \sqrt{\frac{2 G M_{E}}{R_{E}}} = \sqrt{2 g R_{E}} (∵ g = \frac{G M_{E}}{R_{E}^{2}})$

PHXI08:GRAVITATION

359816 There is no atmosphere on moon because

1 It revolves round the earth

2 It is closer to earth

3 The escape velocity of the gas molecules is less than their root mean square velocity here

4 None of the above

Explanation:

As, the escape velocity on the surface of moon's is small i.e., 2.4 $k m / s$ in comparision to 11.2 $k m / s$ that of earth and also this velocity is smaller than $r m s$ velocity of gas molecules, therefore gas molecules escape moon's gravitation. Thus, moon cannot have atmosphere.

PHXI08:GRAVITATION

359817 A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energies is

1 Positive

2 Negative

3 Zero

4 May be positive or negative depending upon its initial velocity

Explanation:

When a body is projected with escape velocity then $T . E$ is zero at infinity.
$\frac{- G M m}{R} + \frac{1}{2} m v_{e}^{2} = 0$
If the initial velocity is less than $v_{e}$ then the body is in bound with earth and hence $T . E$ $= (-)$ ve

PHXI08:GRAVITATION

359813 The mass of the moon is $\frac{1}{144}$ times the mass ofa planet and its diameter is $\frac{1}{16}$ times the diameter of a planet. If the escape velocity on the planet is $v$ , the escape velocity on the moon will be

1

\frac{v}{4}

2

\frac{v}{12}

3

\frac{v}{3}

4

\frac{v}{6}

Explanation:

Escapee velocity on the planet,
$v = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{4 G M}{D}} (1)$
Where, $M =$ Mass of the planet
$R =$ Radius of the planet
$G =$ Universal gravitational constant
$D =$ Diameter of the planet
$v \propto \sqrt{\frac{M}{D}}$
For moon, $v^{'} = \sqrt{\frac{2 G M^{'}}{R^{'}}} = \sqrt{\frac{4 G M^{'}}{D^{'}}} (2)$
where, $M^{'} =$ mass of the moon
$R^{'} =$ Radius of the moon;
$D^{'} =$ diameter of the moon
Given : $M^{'} = \frac{M}{144}$ and $D^{'} = \frac{D}{16}$
From equation (1) and (2), we get
$\frac{v^{'}}{v} = \sqrt{\frac{M^{'} D}{D^{'} M}}; \frac{v^{'}}{v} = \sqrt{\frac{\frac{M}{\frac{144}{D}} \times \frac{D}{1}}{16} \times \frac{M}{1}}$
$= \frac{1}{3} \Rightarrow v^{'} = \frac{v}{3}$

JEE - 2024

PHXI08:GRAVITATION

359814 A stationary object is released from a point $P$ at a distance $3 R$ from the centre of the Moon which has radius $R$ and Mass $M$ . Which one of the following expressions gives the speed of the object on hitting the Moon?
supporting img

1

{(\frac{2 G M}{3 R})}^{1 / 2}

2

{(\frac{4 G M}{3 R})}^{1 / 2}

3

{(\frac{2 G M}{R})}^{1 / 2}

4

{(\frac{G M}{R})}^{1 / 2}

Explanation:

$U_{i} + K_{i} = U_{f} + K_{f}$
$\begin{aligned} \frac{- G M m}{3 R} + 0 = \frac{- G M m}{R} + \frac{1}{2} m v_{f}^{2} \\ \Rightarrow \frac{1}{2} m v_{f}^{2} = \frac{G M m}{R} - \frac{G M m}{3 R} \end{aligned}$
$\frac{1}{2} v_{f}^{2} = \frac{2}{3} \frac{G M}{R} \Rightarrow v_{f} = {(\frac{4}{3} \frac{G M}{R})}^{\frac{1}{2}}$

PHXI08:GRAVITATION

359815 The escape velocity from the surface of the Earth is (where $R_{E}$ is the radius of the earth)

1

\sqrt{2 g R_{E}}

2

\sqrt{g R_{E}}

3

2 \sqrt{g R_{E}}

4

\sqrt{3 g R_{E}}

Explanation:

The escape velocity form the surface of the Earth is
$v_{e} = \sqrt{\frac{2 G M_{E}}{R_{E}}} = \sqrt{2 g R_{E}} (∵ g = \frac{G M_{E}}{R_{E}^{2}})$

PHXI08:GRAVITATION

359816 There is no atmosphere on moon because

1 It revolves round the earth

2 It is closer to earth

3 The escape velocity of the gas molecules is less than their root mean square velocity here

4 None of the above

Explanation:

As, the escape velocity on the surface of moon's is small i.e., 2.4 $k m / s$ in comparision to 11.2 $k m / s$ that of earth and also this velocity is smaller than $r m s$ velocity of gas molecules, therefore gas molecules escape moon's gravitation. Thus, moon cannot have atmosphere.

PHXI08:GRAVITATION

359817 A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energies is

1 Positive

2 Negative

3 Zero

4 May be positive or negative depending upon its initial velocity

Explanation:

When a body is projected with escape velocity then $T . E$ is zero at infinity.
$\frac{- G M m}{R} + \frac{1}{2} m v_{e}^{2} = 0$
If the initial velocity is less than $v_{e}$ then the body is in bound with earth and hence $T . E$ $= (-)$ ve