359813
The mass of the moon is \(\dfrac{1}{144}\) times the mass ofa planet and its diameter is \(\dfrac{1}{16}\) times the diameter of a planet. If the escape velocity on the planet is \(v\), the escape velocity on the moon will be
1 \(\dfrac{v}{4}\)
2 \(\dfrac{v}{12}\)
3 \(\dfrac{v}{3}\)
4 \(\dfrac{v}{6}\)
Explanation:
Escapee velocity on the planet, \(v = \sqrt {\frac{{2GM}}{R}} = \sqrt {\frac{{4GM}}{D}} \quad \quad (1)\) Where, \(M=\) Mass of the planet \(R=\) Radius of the planet \(G=\) Universal gravitational constant \(D=\) Diameter of the planet \(v \propto \sqrt{\dfrac{M}{D}}\) For moon, \(v' = \sqrt {\frac{{2GM'}}{{R'}}} = \sqrt {\frac{{4GM'}}{{D'}}} \quad \quad (2)\) where, \(M^{\prime}=\) mass of the moon \(R^{\prime}=\) Radius of the moon; \(D^{\prime}=\) diameter of the moon Given : \(M^{\prime}=\dfrac{M}{144}\) and \(D^{\prime}=\dfrac{D}{16}\) From equation (1) and (2), we get \(\frac{{v'}}{v} = \sqrt {\frac{{M'D}}{{D'M}}} ;\frac{{v'}}{v} = \sqrt {\frac{{\frac{M}{{\frac{{144}}{D}}} \times \frac{D}{1}}}{{16}} \times \frac{M}{1}} \) \(=\dfrac{1}{3} \Rightarrow v^{\prime}=\dfrac{v}{3}\)
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?
1 \(\left(\dfrac{2 G M}{3 R}\right)^{1 / 2}\)
2 \(\left(\dfrac{4 G M}{3 R}\right)^{1 / 2}\)
3 \(\left(\dfrac{2 G M}{R}\right)^{1 / 2}\)
4 \(\left(\dfrac{G M}{R}\right)^{1 / 2}\)
Explanation:
\(U_{i}+K_{i}=U_{f}+K_{f}\) \(\begin{aligned}& \dfrac{-G M m}{3 R}+0=\dfrac{-G M m}{R}+\dfrac{1}{2} m v_{f}^{2} \\& \Rightarrow \dfrac{1}{2} m v_{f}^{2}=\dfrac{G M m}{R}-\dfrac{G M m}{3 R}\end{aligned}\) \(\dfrac{1}{2} v_{f}^{2}=\dfrac{2}{3} \dfrac{G M}{R} \Rightarrow v_{f}=\left(\dfrac{4}{3} \dfrac{G M}{R}\right)^{\frac{1}{2}}\)
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 \(v_{e}=\sqrt{\dfrac{2 G M_{E}}{R_{E}}}=\sqrt{2 g R_{E}}\left(\because g=\dfrac{G M_{E}}{R_{E}^{2}}\right)\)
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 \(km/s\) in comparision to 11.2 \(km/s\) that of earth and also this velocity is smaller than \(rms\) 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. \(\dfrac{-G M m}{R}+\dfrac{1}{2} m v_{e}^{2}=0\) If the initial velocity is less than \(v_{e}\) then the body is in bound with earth and hence \(T.E\) \(=(-)\) ve
359813
The mass of the moon is \(\dfrac{1}{144}\) times the mass ofa planet and its diameter is \(\dfrac{1}{16}\) times the diameter of a planet. If the escape velocity on the planet is \(v\), the escape velocity on the moon will be
1 \(\dfrac{v}{4}\)
2 \(\dfrac{v}{12}\)
3 \(\dfrac{v}{3}\)
4 \(\dfrac{v}{6}\)
Explanation:
Escapee velocity on the planet, \(v = \sqrt {\frac{{2GM}}{R}} = \sqrt {\frac{{4GM}}{D}} \quad \quad (1)\) Where, \(M=\) Mass of the planet \(R=\) Radius of the planet \(G=\) Universal gravitational constant \(D=\) Diameter of the planet \(v \propto \sqrt{\dfrac{M}{D}}\) For moon, \(v' = \sqrt {\frac{{2GM'}}{{R'}}} = \sqrt {\frac{{4GM'}}{{D'}}} \quad \quad (2)\) where, \(M^{\prime}=\) mass of the moon \(R^{\prime}=\) Radius of the moon; \(D^{\prime}=\) diameter of the moon Given : \(M^{\prime}=\dfrac{M}{144}\) and \(D^{\prime}=\dfrac{D}{16}\) From equation (1) and (2), we get \(\frac{{v'}}{v} = \sqrt {\frac{{M'D}}{{D'M}}} ;\frac{{v'}}{v} = \sqrt {\frac{{\frac{M}{{\frac{{144}}{D}}} \times \frac{D}{1}}}{{16}} \times \frac{M}{1}} \) \(=\dfrac{1}{3} \Rightarrow v^{\prime}=\dfrac{v}{3}\)
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?
1 \(\left(\dfrac{2 G M}{3 R}\right)^{1 / 2}\)
2 \(\left(\dfrac{4 G M}{3 R}\right)^{1 / 2}\)
3 \(\left(\dfrac{2 G M}{R}\right)^{1 / 2}\)
4 \(\left(\dfrac{G M}{R}\right)^{1 / 2}\)
Explanation:
\(U_{i}+K_{i}=U_{f}+K_{f}\) \(\begin{aligned}& \dfrac{-G M m}{3 R}+0=\dfrac{-G M m}{R}+\dfrac{1}{2} m v_{f}^{2} \\& \Rightarrow \dfrac{1}{2} m v_{f}^{2}=\dfrac{G M m}{R}-\dfrac{G M m}{3 R}\end{aligned}\) \(\dfrac{1}{2} v_{f}^{2}=\dfrac{2}{3} \dfrac{G M}{R} \Rightarrow v_{f}=\left(\dfrac{4}{3} \dfrac{G M}{R}\right)^{\frac{1}{2}}\)
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 \(v_{e}=\sqrt{\dfrac{2 G M_{E}}{R_{E}}}=\sqrt{2 g R_{E}}\left(\because g=\dfrac{G M_{E}}{R_{E}^{2}}\right)\)
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 \(km/s\) in comparision to 11.2 \(km/s\) that of earth and also this velocity is smaller than \(rms\) 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. \(\dfrac{-G M m}{R}+\dfrac{1}{2} m v_{e}^{2}=0\) If the initial velocity is less than \(v_{e}\) then the body is in bound with earth and hence \(T.E\) \(=(-)\) ve
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PHXI08:GRAVITATION
359813
The mass of the moon is \(\dfrac{1}{144}\) times the mass ofa planet and its diameter is \(\dfrac{1}{16}\) times the diameter of a planet. If the escape velocity on the planet is \(v\), the escape velocity on the moon will be
1 \(\dfrac{v}{4}\)
2 \(\dfrac{v}{12}\)
3 \(\dfrac{v}{3}\)
4 \(\dfrac{v}{6}\)
Explanation:
Escapee velocity on the planet, \(v = \sqrt {\frac{{2GM}}{R}} = \sqrt {\frac{{4GM}}{D}} \quad \quad (1)\) Where, \(M=\) Mass of the planet \(R=\) Radius of the planet \(G=\) Universal gravitational constant \(D=\) Diameter of the planet \(v \propto \sqrt{\dfrac{M}{D}}\) For moon, \(v' = \sqrt {\frac{{2GM'}}{{R'}}} = \sqrt {\frac{{4GM'}}{{D'}}} \quad \quad (2)\) where, \(M^{\prime}=\) mass of the moon \(R^{\prime}=\) Radius of the moon; \(D^{\prime}=\) diameter of the moon Given : \(M^{\prime}=\dfrac{M}{144}\) and \(D^{\prime}=\dfrac{D}{16}\) From equation (1) and (2), we get \(\frac{{v'}}{v} = \sqrt {\frac{{M'D}}{{D'M}}} ;\frac{{v'}}{v} = \sqrt {\frac{{\frac{M}{{\frac{{144}}{D}}} \times \frac{D}{1}}}{{16}} \times \frac{M}{1}} \) \(=\dfrac{1}{3} \Rightarrow v^{\prime}=\dfrac{v}{3}\)
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?
1 \(\left(\dfrac{2 G M}{3 R}\right)^{1 / 2}\)
2 \(\left(\dfrac{4 G M}{3 R}\right)^{1 / 2}\)
3 \(\left(\dfrac{2 G M}{R}\right)^{1 / 2}\)
4 \(\left(\dfrac{G M}{R}\right)^{1 / 2}\)
Explanation:
\(U_{i}+K_{i}=U_{f}+K_{f}\) \(\begin{aligned}& \dfrac{-G M m}{3 R}+0=\dfrac{-G M m}{R}+\dfrac{1}{2} m v_{f}^{2} \\& \Rightarrow \dfrac{1}{2} m v_{f}^{2}=\dfrac{G M m}{R}-\dfrac{G M m}{3 R}\end{aligned}\) \(\dfrac{1}{2} v_{f}^{2}=\dfrac{2}{3} \dfrac{G M}{R} \Rightarrow v_{f}=\left(\dfrac{4}{3} \dfrac{G M}{R}\right)^{\frac{1}{2}}\)
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 \(v_{e}=\sqrt{\dfrac{2 G M_{E}}{R_{E}}}=\sqrt{2 g R_{E}}\left(\because g=\dfrac{G M_{E}}{R_{E}^{2}}\right)\)
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 \(km/s\) in comparision to 11.2 \(km/s\) that of earth and also this velocity is smaller than \(rms\) 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. \(\dfrac{-G M m}{R}+\dfrac{1}{2} m v_{e}^{2}=0\) If the initial velocity is less than \(v_{e}\) then the body is in bound with earth and hence \(T.E\) \(=(-)\) ve
359813
The mass of the moon is \(\dfrac{1}{144}\) times the mass ofa planet and its diameter is \(\dfrac{1}{16}\) times the diameter of a planet. If the escape velocity on the planet is \(v\), the escape velocity on the moon will be
1 \(\dfrac{v}{4}\)
2 \(\dfrac{v}{12}\)
3 \(\dfrac{v}{3}\)
4 \(\dfrac{v}{6}\)
Explanation:
Escapee velocity on the planet, \(v = \sqrt {\frac{{2GM}}{R}} = \sqrt {\frac{{4GM}}{D}} \quad \quad (1)\) Where, \(M=\) Mass of the planet \(R=\) Radius of the planet \(G=\) Universal gravitational constant \(D=\) Diameter of the planet \(v \propto \sqrt{\dfrac{M}{D}}\) For moon, \(v' = \sqrt {\frac{{2GM'}}{{R'}}} = \sqrt {\frac{{4GM'}}{{D'}}} \quad \quad (2)\) where, \(M^{\prime}=\) mass of the moon \(R^{\prime}=\) Radius of the moon; \(D^{\prime}=\) diameter of the moon Given : \(M^{\prime}=\dfrac{M}{144}\) and \(D^{\prime}=\dfrac{D}{16}\) From equation (1) and (2), we get \(\frac{{v'}}{v} = \sqrt {\frac{{M'D}}{{D'M}}} ;\frac{{v'}}{v} = \sqrt {\frac{{\frac{M}{{\frac{{144}}{D}}} \times \frac{D}{1}}}{{16}} \times \frac{M}{1}} \) \(=\dfrac{1}{3} \Rightarrow v^{\prime}=\dfrac{v}{3}\)
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?
1 \(\left(\dfrac{2 G M}{3 R}\right)^{1 / 2}\)
2 \(\left(\dfrac{4 G M}{3 R}\right)^{1 / 2}\)
3 \(\left(\dfrac{2 G M}{R}\right)^{1 / 2}\)
4 \(\left(\dfrac{G M}{R}\right)^{1 / 2}\)
Explanation:
\(U_{i}+K_{i}=U_{f}+K_{f}\) \(\begin{aligned}& \dfrac{-G M m}{3 R}+0=\dfrac{-G M m}{R}+\dfrac{1}{2} m v_{f}^{2} \\& \Rightarrow \dfrac{1}{2} m v_{f}^{2}=\dfrac{G M m}{R}-\dfrac{G M m}{3 R}\end{aligned}\) \(\dfrac{1}{2} v_{f}^{2}=\dfrac{2}{3} \dfrac{G M}{R} \Rightarrow v_{f}=\left(\dfrac{4}{3} \dfrac{G M}{R}\right)^{\frac{1}{2}}\)
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 \(v_{e}=\sqrt{\dfrac{2 G M_{E}}{R_{E}}}=\sqrt{2 g R_{E}}\left(\because g=\dfrac{G M_{E}}{R_{E}^{2}}\right)\)
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 \(km/s\) in comparision to 11.2 \(km/s\) that of earth and also this velocity is smaller than \(rms\) 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. \(\dfrac{-G M m}{R}+\dfrac{1}{2} m v_{e}^{2}=0\) If the initial velocity is less than \(v_{e}\) then the body is in bound with earth and hence \(T.E\) \(=(-)\) ve
359813
The mass of the moon is \(\dfrac{1}{144}\) times the mass ofa planet and its diameter is \(\dfrac{1}{16}\) times the diameter of a planet. If the escape velocity on the planet is \(v\), the escape velocity on the moon will be
1 \(\dfrac{v}{4}\)
2 \(\dfrac{v}{12}\)
3 \(\dfrac{v}{3}\)
4 \(\dfrac{v}{6}\)
Explanation:
Escapee velocity on the planet, \(v = \sqrt {\frac{{2GM}}{R}} = \sqrt {\frac{{4GM}}{D}} \quad \quad (1)\) Where, \(M=\) Mass of the planet \(R=\) Radius of the planet \(G=\) Universal gravitational constant \(D=\) Diameter of the planet \(v \propto \sqrt{\dfrac{M}{D}}\) For moon, \(v' = \sqrt {\frac{{2GM'}}{{R'}}} = \sqrt {\frac{{4GM'}}{{D'}}} \quad \quad (2)\) where, \(M^{\prime}=\) mass of the moon \(R^{\prime}=\) Radius of the moon; \(D^{\prime}=\) diameter of the moon Given : \(M^{\prime}=\dfrac{M}{144}\) and \(D^{\prime}=\dfrac{D}{16}\) From equation (1) and (2), we get \(\frac{{v'}}{v} = \sqrt {\frac{{M'D}}{{D'M}}} ;\frac{{v'}}{v} = \sqrt {\frac{{\frac{M}{{\frac{{144}}{D}}} \times \frac{D}{1}}}{{16}} \times \frac{M}{1}} \) \(=\dfrac{1}{3} \Rightarrow v^{\prime}=\dfrac{v}{3}\)
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?
1 \(\left(\dfrac{2 G M}{3 R}\right)^{1 / 2}\)
2 \(\left(\dfrac{4 G M}{3 R}\right)^{1 / 2}\)
3 \(\left(\dfrac{2 G M}{R}\right)^{1 / 2}\)
4 \(\left(\dfrac{G M}{R}\right)^{1 / 2}\)
Explanation:
\(U_{i}+K_{i}=U_{f}+K_{f}\) \(\begin{aligned}& \dfrac{-G M m}{3 R}+0=\dfrac{-G M m}{R}+\dfrac{1}{2} m v_{f}^{2} \\& \Rightarrow \dfrac{1}{2} m v_{f}^{2}=\dfrac{G M m}{R}-\dfrac{G M m}{3 R}\end{aligned}\) \(\dfrac{1}{2} v_{f}^{2}=\dfrac{2}{3} \dfrac{G M}{R} \Rightarrow v_{f}=\left(\dfrac{4}{3} \dfrac{G M}{R}\right)^{\frac{1}{2}}\)
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 \(v_{e}=\sqrt{\dfrac{2 G M_{E}}{R_{E}}}=\sqrt{2 g R_{E}}\left(\because g=\dfrac{G M_{E}}{R_{E}^{2}}\right)\)
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 \(km/s\) in comparision to 11.2 \(km/s\) that of earth and also this velocity is smaller than \(rms\) 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. \(\dfrac{-G M m}{R}+\dfrac{1}{2} m v_{e}^{2}=0\) If the initial velocity is less than \(v_{e}\) then the body is in bound with earth and hence \(T.E\) \(=(-)\) ve
