138686
The escape velocity from a planet is $V_{\mathrm{e}}$. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be
1 $\mathrm{v}_{\mathrm{e}}$
2 $\mathrm{v}_{\mathrm{e}} / \sqrt{2}$
3 $\mathrm{v}_{\mathrm{e}} / 2$
4 Zero
Explanation:
Gravitational potential at the surface of the earth $\mathrm{V}_{\mathrm{s}}=\frac{-\mathrm{GM}}{\mathrm{R}}$ Gravitational potential at the center of earth $\mathrm{V}_{\mathrm{C}}=\frac{-3 \mathrm{GM}}{2 \mathrm{R}}$ Loss of potential energy of particle $=$ Gain of kinetic energy $\mathrm{m}\left(\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{c}}\right)=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~m}\left(-\frac{\mathrm{GM}}{\mathrm{R}}-\left(-\frac{3 \mathrm{GM}}{2 \mathrm{R}}\right)\right)=\frac{1}{2} \mathrm{mv}^{2}$ $\frac{\mathrm{GM}}{\mathrm{R}}\left(-1+\frac{3}{2}\right)=\frac{1}{2} \mathrm{v}^{2}$ $\frac{1}{2} \frac{\mathrm{GM}}{\mathrm{R}}=\frac{1}{2} \mathrm{v}^{2}$ \(\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}} \quad\left(\begin{array}{l}\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2} \\ \text { and } \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\end{array}\right)\) $\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}^2} \times \mathrm{R}$ $\mathrm{v}^2=\mathrm{gR}$ $\mathrm{v}^2=\frac{\mathrm{v}_{\mathrm{e}}^2}{2}$ $\mathrm{v}=\frac{\mathrm{v}_{\mathrm{e}}}{\sqrt{2}}$
BITSAT-2020
Gravitation
138687
Two planets $A$ and $B$ have the same material density. If the radius of $A$ is twice that of $B$, then the ratio of the escape velocity $v_{A} / v_{B}$ is
1 2
2 $\sqrt{2}$
3 $1 / \sqrt{2}$
4 $1 / 2$
Explanation:
A Given that, $R_{A}=2 R_{B}$ We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ \(\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}} \quad\left(\begin{array}{l}\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2} \\ \text { and } \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\end{array}\right)\) $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{8 \mathrm{G} \pi \mathrm{R}^{2} \rho}{3}}$ $\therefore \quad \mathrm{v}_{\mathrm{e}} \propto \mathrm{R}$ Then, $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{2 \mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=2$
BITSAT-2007
Gravitation
138689
A satellite of mass $m$ is circulating around the earth with constant angular velocity. If the radius of the orbit is $R_{0}$ and mass of the Earth is $M$, the angular momentum of satellite about the centre of Earth is
B We know that, Angular momentum $(\mathrm{L}) =\mathrm{m} \cdot \mathrm{v}_{\mathrm{o}} \cdot \mathrm{R}_{\mathrm{o}}$ $=\text { mass } \times \text { orbital velocity } \times \text { radius }$ $=\mathrm{m} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}_{0}}} \times \mathrm{R}_{0}$ $=\mathrm{m} \sqrt{\mathrm{GMR}_{0}}$
BITSAT-2016
Gravitation
138690
The acceleration due to gravity on the surface of the moon is $1 / 6$ that on the surface of earth and the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and moon will be
1 $\frac{\sqrt{6}}{2}$
2 $\sqrt{24}$
3 3
4 $\frac{\sqrt{3}}{2}$
Explanation:
B Given that, $\mathrm{g}_{\text {moon }}=\frac{1}{6}$ of earth $\mathrm{R}_{\text {moon }}=\frac{1}{4}$ of earth We know that, $\mathrm{v}_{\text {escape }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{2 \mathrm{gr}}$ $\frac{\mathrm{v}_{\text {earth }}}{\mathrm{v}_{\text {moon }}}=\sqrt{\frac{\mathrm{g}_{\text {earth }}}{\mathrm{g}_{\text {moon }}} \times \frac{\mathrm{R}_{\text {earth }}}{\mathrm{R}_{\text {moon }}}}$ $\frac{\mathrm{v}_{\text {earth }}}{\mathrm{v}_{\text {moon }}}=\sqrt{6 \times 4}=\sqrt{24}$
138686
The escape velocity from a planet is $V_{\mathrm{e}}$. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be
1 $\mathrm{v}_{\mathrm{e}}$
2 $\mathrm{v}_{\mathrm{e}} / \sqrt{2}$
3 $\mathrm{v}_{\mathrm{e}} / 2$
4 Zero
Explanation:
Gravitational potential at the surface of the earth $\mathrm{V}_{\mathrm{s}}=\frac{-\mathrm{GM}}{\mathrm{R}}$ Gravitational potential at the center of earth $\mathrm{V}_{\mathrm{C}}=\frac{-3 \mathrm{GM}}{2 \mathrm{R}}$ Loss of potential energy of particle $=$ Gain of kinetic energy $\mathrm{m}\left(\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{c}}\right)=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~m}\left(-\frac{\mathrm{GM}}{\mathrm{R}}-\left(-\frac{3 \mathrm{GM}}{2 \mathrm{R}}\right)\right)=\frac{1}{2} \mathrm{mv}^{2}$ $\frac{\mathrm{GM}}{\mathrm{R}}\left(-1+\frac{3}{2}\right)=\frac{1}{2} \mathrm{v}^{2}$ $\frac{1}{2} \frac{\mathrm{GM}}{\mathrm{R}}=\frac{1}{2} \mathrm{v}^{2}$ \(\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}} \quad\left(\begin{array}{l}\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2} \\ \text { and } \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\end{array}\right)\) $\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}^2} \times \mathrm{R}$ $\mathrm{v}^2=\mathrm{gR}$ $\mathrm{v}^2=\frac{\mathrm{v}_{\mathrm{e}}^2}{2}$ $\mathrm{v}=\frac{\mathrm{v}_{\mathrm{e}}}{\sqrt{2}}$
BITSAT-2020
Gravitation
138687
Two planets $A$ and $B$ have the same material density. If the radius of $A$ is twice that of $B$, then the ratio of the escape velocity $v_{A} / v_{B}$ is
1 2
2 $\sqrt{2}$
3 $1 / \sqrt{2}$
4 $1 / 2$
Explanation:
A Given that, $R_{A}=2 R_{B}$ We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ \(\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}} \quad\left(\begin{array}{l}\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2} \\ \text { and } \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\end{array}\right)\) $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{8 \mathrm{G} \pi \mathrm{R}^{2} \rho}{3}}$ $\therefore \quad \mathrm{v}_{\mathrm{e}} \propto \mathrm{R}$ Then, $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{2 \mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=2$
BITSAT-2007
Gravitation
138689
A satellite of mass $m$ is circulating around the earth with constant angular velocity. If the radius of the orbit is $R_{0}$ and mass of the Earth is $M$, the angular momentum of satellite about the centre of Earth is
B We know that, Angular momentum $(\mathrm{L}) =\mathrm{m} \cdot \mathrm{v}_{\mathrm{o}} \cdot \mathrm{R}_{\mathrm{o}}$ $=\text { mass } \times \text { orbital velocity } \times \text { radius }$ $=\mathrm{m} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}_{0}}} \times \mathrm{R}_{0}$ $=\mathrm{m} \sqrt{\mathrm{GMR}_{0}}$
BITSAT-2016
Gravitation
138690
The acceleration due to gravity on the surface of the moon is $1 / 6$ that on the surface of earth and the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and moon will be
1 $\frac{\sqrt{6}}{2}$
2 $\sqrt{24}$
3 3
4 $\frac{\sqrt{3}}{2}$
Explanation:
B Given that, $\mathrm{g}_{\text {moon }}=\frac{1}{6}$ of earth $\mathrm{R}_{\text {moon }}=\frac{1}{4}$ of earth We know that, $\mathrm{v}_{\text {escape }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{2 \mathrm{gr}}$ $\frac{\mathrm{v}_{\text {earth }}}{\mathrm{v}_{\text {moon }}}=\sqrt{\frac{\mathrm{g}_{\text {earth }}}{\mathrm{g}_{\text {moon }}} \times \frac{\mathrm{R}_{\text {earth }}}{\mathrm{R}_{\text {moon }}}}$ $\frac{\mathrm{v}_{\text {earth }}}{\mathrm{v}_{\text {moon }}}=\sqrt{6 \times 4}=\sqrt{24}$
138686
The escape velocity from a planet is $V_{\mathrm{e}}$. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be
1 $\mathrm{v}_{\mathrm{e}}$
2 $\mathrm{v}_{\mathrm{e}} / \sqrt{2}$
3 $\mathrm{v}_{\mathrm{e}} / 2$
4 Zero
Explanation:
Gravitational potential at the surface of the earth $\mathrm{V}_{\mathrm{s}}=\frac{-\mathrm{GM}}{\mathrm{R}}$ Gravitational potential at the center of earth $\mathrm{V}_{\mathrm{C}}=\frac{-3 \mathrm{GM}}{2 \mathrm{R}}$ Loss of potential energy of particle $=$ Gain of kinetic energy $\mathrm{m}\left(\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{c}}\right)=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~m}\left(-\frac{\mathrm{GM}}{\mathrm{R}}-\left(-\frac{3 \mathrm{GM}}{2 \mathrm{R}}\right)\right)=\frac{1}{2} \mathrm{mv}^{2}$ $\frac{\mathrm{GM}}{\mathrm{R}}\left(-1+\frac{3}{2}\right)=\frac{1}{2} \mathrm{v}^{2}$ $\frac{1}{2} \frac{\mathrm{GM}}{\mathrm{R}}=\frac{1}{2} \mathrm{v}^{2}$ \(\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}} \quad\left(\begin{array}{l}\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2} \\ \text { and } \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\end{array}\right)\) $\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}^2} \times \mathrm{R}$ $\mathrm{v}^2=\mathrm{gR}$ $\mathrm{v}^2=\frac{\mathrm{v}_{\mathrm{e}}^2}{2}$ $\mathrm{v}=\frac{\mathrm{v}_{\mathrm{e}}}{\sqrt{2}}$
BITSAT-2020
Gravitation
138687
Two planets $A$ and $B$ have the same material density. If the radius of $A$ is twice that of $B$, then the ratio of the escape velocity $v_{A} / v_{B}$ is
1 2
2 $\sqrt{2}$
3 $1 / \sqrt{2}$
4 $1 / 2$
Explanation:
A Given that, $R_{A}=2 R_{B}$ We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ \(\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}} \quad\left(\begin{array}{l}\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2} \\ \text { and } \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\end{array}\right)\) $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{8 \mathrm{G} \pi \mathrm{R}^{2} \rho}{3}}$ $\therefore \quad \mathrm{v}_{\mathrm{e}} \propto \mathrm{R}$ Then, $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{2 \mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=2$
BITSAT-2007
Gravitation
138689
A satellite of mass $m$ is circulating around the earth with constant angular velocity. If the radius of the orbit is $R_{0}$ and mass of the Earth is $M$, the angular momentum of satellite about the centre of Earth is
B We know that, Angular momentum $(\mathrm{L}) =\mathrm{m} \cdot \mathrm{v}_{\mathrm{o}} \cdot \mathrm{R}_{\mathrm{o}}$ $=\text { mass } \times \text { orbital velocity } \times \text { radius }$ $=\mathrm{m} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}_{0}}} \times \mathrm{R}_{0}$ $=\mathrm{m} \sqrt{\mathrm{GMR}_{0}}$
BITSAT-2016
Gravitation
138690
The acceleration due to gravity on the surface of the moon is $1 / 6$ that on the surface of earth and the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and moon will be
1 $\frac{\sqrt{6}}{2}$
2 $\sqrt{24}$
3 3
4 $\frac{\sqrt{3}}{2}$
Explanation:
B Given that, $\mathrm{g}_{\text {moon }}=\frac{1}{6}$ of earth $\mathrm{R}_{\text {moon }}=\frac{1}{4}$ of earth We know that, $\mathrm{v}_{\text {escape }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{2 \mathrm{gr}}$ $\frac{\mathrm{v}_{\text {earth }}}{\mathrm{v}_{\text {moon }}}=\sqrt{\frac{\mathrm{g}_{\text {earth }}}{\mathrm{g}_{\text {moon }}} \times \frac{\mathrm{R}_{\text {earth }}}{\mathrm{R}_{\text {moon }}}}$ $\frac{\mathrm{v}_{\text {earth }}}{\mathrm{v}_{\text {moon }}}=\sqrt{6 \times 4}=\sqrt{24}$
138686
The escape velocity from a planet is $V_{\mathrm{e}}$. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be
1 $\mathrm{v}_{\mathrm{e}}$
2 $\mathrm{v}_{\mathrm{e}} / \sqrt{2}$
3 $\mathrm{v}_{\mathrm{e}} / 2$
4 Zero
Explanation:
Gravitational potential at the surface of the earth $\mathrm{V}_{\mathrm{s}}=\frac{-\mathrm{GM}}{\mathrm{R}}$ Gravitational potential at the center of earth $\mathrm{V}_{\mathrm{C}}=\frac{-3 \mathrm{GM}}{2 \mathrm{R}}$ Loss of potential energy of particle $=$ Gain of kinetic energy $\mathrm{m}\left(\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{c}}\right)=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~m}\left(-\frac{\mathrm{GM}}{\mathrm{R}}-\left(-\frac{3 \mathrm{GM}}{2 \mathrm{R}}\right)\right)=\frac{1}{2} \mathrm{mv}^{2}$ $\frac{\mathrm{GM}}{\mathrm{R}}\left(-1+\frac{3}{2}\right)=\frac{1}{2} \mathrm{v}^{2}$ $\frac{1}{2} \frac{\mathrm{GM}}{\mathrm{R}}=\frac{1}{2} \mathrm{v}^{2}$ \(\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}} \quad\left(\begin{array}{l}\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2} \\ \text { and } \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\end{array}\right)\) $\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}^2} \times \mathrm{R}$ $\mathrm{v}^2=\mathrm{gR}$ $\mathrm{v}^2=\frac{\mathrm{v}_{\mathrm{e}}^2}{2}$ $\mathrm{v}=\frac{\mathrm{v}_{\mathrm{e}}}{\sqrt{2}}$
BITSAT-2020
Gravitation
138687
Two planets $A$ and $B$ have the same material density. If the radius of $A$ is twice that of $B$, then the ratio of the escape velocity $v_{A} / v_{B}$ is
1 2
2 $\sqrt{2}$
3 $1 / \sqrt{2}$
4 $1 / 2$
Explanation:
A Given that, $R_{A}=2 R_{B}$ We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ \(\mathrm{v}^2=\frac{\mathrm{GM}}{\mathrm{R}} \quad\left(\begin{array}{l}\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2} \\ \text { and } \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\end{array}\right)\) $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{8 \mathrm{G} \pi \mathrm{R}^{2} \rho}{3}}$ $\therefore \quad \mathrm{v}_{\mathrm{e}} \propto \mathrm{R}$ Then, $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{2 \mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=2$
BITSAT-2007
Gravitation
138689
A satellite of mass $m$ is circulating around the earth with constant angular velocity. If the radius of the orbit is $R_{0}$ and mass of the Earth is $M$, the angular momentum of satellite about the centre of Earth is
B We know that, Angular momentum $(\mathrm{L}) =\mathrm{m} \cdot \mathrm{v}_{\mathrm{o}} \cdot \mathrm{R}_{\mathrm{o}}$ $=\text { mass } \times \text { orbital velocity } \times \text { radius }$ $=\mathrm{m} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}_{0}}} \times \mathrm{R}_{0}$ $=\mathrm{m} \sqrt{\mathrm{GMR}_{0}}$
BITSAT-2016
Gravitation
138690
The acceleration due to gravity on the surface of the moon is $1 / 6$ that on the surface of earth and the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and moon will be
1 $\frac{\sqrt{6}}{2}$
2 $\sqrt{24}$
3 3
4 $\frac{\sqrt{3}}{2}$
Explanation:
B Given that, $\mathrm{g}_{\text {moon }}=\frac{1}{6}$ of earth $\mathrm{R}_{\text {moon }}=\frac{1}{4}$ of earth We know that, $\mathrm{v}_{\text {escape }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{2 \mathrm{gr}}$ $\frac{\mathrm{v}_{\text {earth }}}{\mathrm{v}_{\text {moon }}}=\sqrt{\frac{\mathrm{g}_{\text {earth }}}{\mathrm{g}_{\text {moon }}} \times \frac{\mathrm{R}_{\text {earth }}}{\mathrm{R}_{\text {moon }}}}$ $\frac{\mathrm{v}_{\text {earth }}}{\mathrm{v}_{\text {moon }}}=\sqrt{6 \times 4}=\sqrt{24}$
