D Given that, Radius of a planet $=\mathrm{R}$ Density of planet $=\rho$ Escape velocity on its surface $\left(\mathrm{v}_{\mathrm{e}}\right)=$ ? We know that, Escape velocity \(\left(v_e\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \quad\left(\begin{array}{c}\mathrm{M}=\mathrm{V} \times \rho \\ =\frac{4}{3} \pi \mathrm{R}^3 \times \rho\end{array}\right)\) $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\mathrm{R} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho}$
BCECE-2011
Gravitation
138736
If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth, the height of the satellite above the surface of the earth is-
1 $2 \mathrm{R}$
2 $\frac{\mathrm{R}}{2}$
3 $\mathrm{R}$
4 $\frac{\mathrm{R}}{4}$
Explanation:
C According to question, Orbital velocity, $\mathrm{v}_{\mathrm{o}}=\frac{\mathrm{v}_{\mathrm{e}}}{2}=\frac{1}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \quad\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\right)$ $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}} \ldots . . \text { (i) }$ We know that, orbital velocity at height $\mathrm{h}$, $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}} .$ On comparing equation (i) and (ii) - $\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}}$ Squaring both sides, $\frac{\mathrm{GM}}{2 \mathrm{R}}=\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}$ So, $\mathrm{R}+\mathrm{h}=2 \mathrm{R}$ $\mathrm{h}=2 \mathrm{R}-\mathrm{R}$ $\mathrm{h}=\mathrm{R}$
BCECE-2011
Gravitation
138737
The escape velocity of a projectile on the earth's surface is $11.2 \mathrm{kms}^{-1}$. A body is projected out with thrice this speed. The speed of the body far away from the earth will be :
1 $22.4 \mathrm{kms}^{-1}$
2 $31.7 \mathrm{kms}^{-1}$
3 $33.6 \mathrm{kms}^{-1}$
4 none of these
Explanation:
B Given that, $\mathrm{v}_{\mathrm{e}}=11.2 \mathrm{kms}^{-1}$ $\mathrm{v}_{\mathrm{P}}=3 \times \mathrm{v}_{\mathrm{e}}$ Mass of body $=\mathrm{m}$ Let velocity of the projectile far away of the earth $=v_{f}$ $\text { Total energy }=\frac{1}{2} \mathrm{mv}_{\mathrm{P}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Now, From the law of conservation of energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{\left(\mathrm{v}_{\mathrm{p}}\right)^{2}-\left(\mathrm{v}_{\mathrm{e}}\right)^{2}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{(3 \times 11.2)^{2}-(11.2)^{2}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{8} \times 11.2$ $\mathrm{v}_{\mathrm{f}}=31.68 \mathrm{kms}^{-1}$
BHU 2006
Gravitation
138738
A satellite is orbiting around the earth. By what percentage should we increase its velocity, so as to enable it escape away from the earth ?
1 $41.4 \%$
2 $50 \%$
3 $82.8 \%$
4 $100 \%$
Explanation:
A We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gr}}$ Orbital velocity $\left(\mathrm{v}_{\mathrm{o}}\right)=\sqrt{\mathrm{gr}}$ So, the increase in velocity $\Delta \mathrm{v} =\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{\mathrm{o}}$ $=\sqrt{2 \mathrm{gr}}-\sqrt{\mathrm{gr}}$ $\Delta \mathrm{v} =(\sqrt{2}-1) \sqrt{\mathrm{gr}}$ $\Delta \mathrm{v} =0.414 \sqrt{\mathrm{gr}}$ Now, $\frac{\Delta \mathrm{v}}{\mathrm{v}_{0}} \times 100=\frac{\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{o}}} \times 100$ $\frac{\Delta \mathrm{v}}{\mathrm{v}_{0}} \times 100=\frac{0.414 \sqrt{\mathrm{gr}}}{\sqrt{\mathrm{gr}}} \times 100$ $\Delta \mathrm{v}=41.4 \%$
BCECE-2003
Gravitation
138739
A body of mass $m$ is situated on the earth in the gravitational field of sun. For the body to escape from the gravitation pull of the solar system the body must be imparted an escape velocity of (assume earth to be stationary)
D Given that, Radius of a planet $=\mathrm{R}$ Density of planet $=\rho$ Escape velocity on its surface $\left(\mathrm{v}_{\mathrm{e}}\right)=$ ? We know that, Escape velocity \(\left(v_e\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \quad\left(\begin{array}{c}\mathrm{M}=\mathrm{V} \times \rho \\ =\frac{4}{3} \pi \mathrm{R}^3 \times \rho\end{array}\right)\) $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\mathrm{R} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho}$
BCECE-2011
Gravitation
138736
If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth, the height of the satellite above the surface of the earth is-
1 $2 \mathrm{R}$
2 $\frac{\mathrm{R}}{2}$
3 $\mathrm{R}$
4 $\frac{\mathrm{R}}{4}$
Explanation:
C According to question, Orbital velocity, $\mathrm{v}_{\mathrm{o}}=\frac{\mathrm{v}_{\mathrm{e}}}{2}=\frac{1}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \quad\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\right)$ $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}} \ldots . . \text { (i) }$ We know that, orbital velocity at height $\mathrm{h}$, $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}} .$ On comparing equation (i) and (ii) - $\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}}$ Squaring both sides, $\frac{\mathrm{GM}}{2 \mathrm{R}}=\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}$ So, $\mathrm{R}+\mathrm{h}=2 \mathrm{R}$ $\mathrm{h}=2 \mathrm{R}-\mathrm{R}$ $\mathrm{h}=\mathrm{R}$
BCECE-2011
Gravitation
138737
The escape velocity of a projectile on the earth's surface is $11.2 \mathrm{kms}^{-1}$. A body is projected out with thrice this speed. The speed of the body far away from the earth will be :
1 $22.4 \mathrm{kms}^{-1}$
2 $31.7 \mathrm{kms}^{-1}$
3 $33.6 \mathrm{kms}^{-1}$
4 none of these
Explanation:
B Given that, $\mathrm{v}_{\mathrm{e}}=11.2 \mathrm{kms}^{-1}$ $\mathrm{v}_{\mathrm{P}}=3 \times \mathrm{v}_{\mathrm{e}}$ Mass of body $=\mathrm{m}$ Let velocity of the projectile far away of the earth $=v_{f}$ $\text { Total energy }=\frac{1}{2} \mathrm{mv}_{\mathrm{P}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Now, From the law of conservation of energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{\left(\mathrm{v}_{\mathrm{p}}\right)^{2}-\left(\mathrm{v}_{\mathrm{e}}\right)^{2}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{(3 \times 11.2)^{2}-(11.2)^{2}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{8} \times 11.2$ $\mathrm{v}_{\mathrm{f}}=31.68 \mathrm{kms}^{-1}$
BHU 2006
Gravitation
138738
A satellite is orbiting around the earth. By what percentage should we increase its velocity, so as to enable it escape away from the earth ?
1 $41.4 \%$
2 $50 \%$
3 $82.8 \%$
4 $100 \%$
Explanation:
A We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gr}}$ Orbital velocity $\left(\mathrm{v}_{\mathrm{o}}\right)=\sqrt{\mathrm{gr}}$ So, the increase in velocity $\Delta \mathrm{v} =\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{\mathrm{o}}$ $=\sqrt{2 \mathrm{gr}}-\sqrt{\mathrm{gr}}$ $\Delta \mathrm{v} =(\sqrt{2}-1) \sqrt{\mathrm{gr}}$ $\Delta \mathrm{v} =0.414 \sqrt{\mathrm{gr}}$ Now, $\frac{\Delta \mathrm{v}}{\mathrm{v}_{0}} \times 100=\frac{\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{o}}} \times 100$ $\frac{\Delta \mathrm{v}}{\mathrm{v}_{0}} \times 100=\frac{0.414 \sqrt{\mathrm{gr}}}{\sqrt{\mathrm{gr}}} \times 100$ $\Delta \mathrm{v}=41.4 \%$
BCECE-2003
Gravitation
138739
A body of mass $m$ is situated on the earth in the gravitational field of sun. For the body to escape from the gravitation pull of the solar system the body must be imparted an escape velocity of (assume earth to be stationary)
D Given that, Radius of a planet $=\mathrm{R}$ Density of planet $=\rho$ Escape velocity on its surface $\left(\mathrm{v}_{\mathrm{e}}\right)=$ ? We know that, Escape velocity \(\left(v_e\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \quad\left(\begin{array}{c}\mathrm{M}=\mathrm{V} \times \rho \\ =\frac{4}{3} \pi \mathrm{R}^3 \times \rho\end{array}\right)\) $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\mathrm{R} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho}$
BCECE-2011
Gravitation
138736
If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth, the height of the satellite above the surface of the earth is-
1 $2 \mathrm{R}$
2 $\frac{\mathrm{R}}{2}$
3 $\mathrm{R}$
4 $\frac{\mathrm{R}}{4}$
Explanation:
C According to question, Orbital velocity, $\mathrm{v}_{\mathrm{o}}=\frac{\mathrm{v}_{\mathrm{e}}}{2}=\frac{1}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \quad\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\right)$ $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}} \ldots . . \text { (i) }$ We know that, orbital velocity at height $\mathrm{h}$, $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}} .$ On comparing equation (i) and (ii) - $\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}}$ Squaring both sides, $\frac{\mathrm{GM}}{2 \mathrm{R}}=\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}$ So, $\mathrm{R}+\mathrm{h}=2 \mathrm{R}$ $\mathrm{h}=2 \mathrm{R}-\mathrm{R}$ $\mathrm{h}=\mathrm{R}$
BCECE-2011
Gravitation
138737
The escape velocity of a projectile on the earth's surface is $11.2 \mathrm{kms}^{-1}$. A body is projected out with thrice this speed. The speed of the body far away from the earth will be :
1 $22.4 \mathrm{kms}^{-1}$
2 $31.7 \mathrm{kms}^{-1}$
3 $33.6 \mathrm{kms}^{-1}$
4 none of these
Explanation:
B Given that, $\mathrm{v}_{\mathrm{e}}=11.2 \mathrm{kms}^{-1}$ $\mathrm{v}_{\mathrm{P}}=3 \times \mathrm{v}_{\mathrm{e}}$ Mass of body $=\mathrm{m}$ Let velocity of the projectile far away of the earth $=v_{f}$ $\text { Total energy }=\frac{1}{2} \mathrm{mv}_{\mathrm{P}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Now, From the law of conservation of energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{\left(\mathrm{v}_{\mathrm{p}}\right)^{2}-\left(\mathrm{v}_{\mathrm{e}}\right)^{2}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{(3 \times 11.2)^{2}-(11.2)^{2}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{8} \times 11.2$ $\mathrm{v}_{\mathrm{f}}=31.68 \mathrm{kms}^{-1}$
BHU 2006
Gravitation
138738
A satellite is orbiting around the earth. By what percentage should we increase its velocity, so as to enable it escape away from the earth ?
1 $41.4 \%$
2 $50 \%$
3 $82.8 \%$
4 $100 \%$
Explanation:
A We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gr}}$ Orbital velocity $\left(\mathrm{v}_{\mathrm{o}}\right)=\sqrt{\mathrm{gr}}$ So, the increase in velocity $\Delta \mathrm{v} =\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{\mathrm{o}}$ $=\sqrt{2 \mathrm{gr}}-\sqrt{\mathrm{gr}}$ $\Delta \mathrm{v} =(\sqrt{2}-1) \sqrt{\mathrm{gr}}$ $\Delta \mathrm{v} =0.414 \sqrt{\mathrm{gr}}$ Now, $\frac{\Delta \mathrm{v}}{\mathrm{v}_{0}} \times 100=\frac{\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{o}}} \times 100$ $\frac{\Delta \mathrm{v}}{\mathrm{v}_{0}} \times 100=\frac{0.414 \sqrt{\mathrm{gr}}}{\sqrt{\mathrm{gr}}} \times 100$ $\Delta \mathrm{v}=41.4 \%$
BCECE-2003
Gravitation
138739
A body of mass $m$ is situated on the earth in the gravitational field of sun. For the body to escape from the gravitation pull of the solar system the body must be imparted an escape velocity of (assume earth to be stationary)
D Given that, Radius of a planet $=\mathrm{R}$ Density of planet $=\rho$ Escape velocity on its surface $\left(\mathrm{v}_{\mathrm{e}}\right)=$ ? We know that, Escape velocity \(\left(v_e\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \quad\left(\begin{array}{c}\mathrm{M}=\mathrm{V} \times \rho \\ =\frac{4}{3} \pi \mathrm{R}^3 \times \rho\end{array}\right)\) $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\mathrm{R} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho}$
BCECE-2011
Gravitation
138736
If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth, the height of the satellite above the surface of the earth is-
1 $2 \mathrm{R}$
2 $\frac{\mathrm{R}}{2}$
3 $\mathrm{R}$
4 $\frac{\mathrm{R}}{4}$
Explanation:
C According to question, Orbital velocity, $\mathrm{v}_{\mathrm{o}}=\frac{\mathrm{v}_{\mathrm{e}}}{2}=\frac{1}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \quad\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\right)$ $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}} \ldots . . \text { (i) }$ We know that, orbital velocity at height $\mathrm{h}$, $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}} .$ On comparing equation (i) and (ii) - $\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}}$ Squaring both sides, $\frac{\mathrm{GM}}{2 \mathrm{R}}=\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}$ So, $\mathrm{R}+\mathrm{h}=2 \mathrm{R}$ $\mathrm{h}=2 \mathrm{R}-\mathrm{R}$ $\mathrm{h}=\mathrm{R}$
BCECE-2011
Gravitation
138737
The escape velocity of a projectile on the earth's surface is $11.2 \mathrm{kms}^{-1}$. A body is projected out with thrice this speed. The speed of the body far away from the earth will be :
1 $22.4 \mathrm{kms}^{-1}$
2 $31.7 \mathrm{kms}^{-1}$
3 $33.6 \mathrm{kms}^{-1}$
4 none of these
Explanation:
B Given that, $\mathrm{v}_{\mathrm{e}}=11.2 \mathrm{kms}^{-1}$ $\mathrm{v}_{\mathrm{P}}=3 \times \mathrm{v}_{\mathrm{e}}$ Mass of body $=\mathrm{m}$ Let velocity of the projectile far away of the earth $=v_{f}$ $\text { Total energy }=\frac{1}{2} \mathrm{mv}_{\mathrm{P}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Now, From the law of conservation of energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{\left(\mathrm{v}_{\mathrm{p}}\right)^{2}-\left(\mathrm{v}_{\mathrm{e}}\right)^{2}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{(3 \times 11.2)^{2}-(11.2)^{2}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{8} \times 11.2$ $\mathrm{v}_{\mathrm{f}}=31.68 \mathrm{kms}^{-1}$
BHU 2006
Gravitation
138738
A satellite is orbiting around the earth. By what percentage should we increase its velocity, so as to enable it escape away from the earth ?
1 $41.4 \%$
2 $50 \%$
3 $82.8 \%$
4 $100 \%$
Explanation:
A We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gr}}$ Orbital velocity $\left(\mathrm{v}_{\mathrm{o}}\right)=\sqrt{\mathrm{gr}}$ So, the increase in velocity $\Delta \mathrm{v} =\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{\mathrm{o}}$ $=\sqrt{2 \mathrm{gr}}-\sqrt{\mathrm{gr}}$ $\Delta \mathrm{v} =(\sqrt{2}-1) \sqrt{\mathrm{gr}}$ $\Delta \mathrm{v} =0.414 \sqrt{\mathrm{gr}}$ Now, $\frac{\Delta \mathrm{v}}{\mathrm{v}_{0}} \times 100=\frac{\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{o}}} \times 100$ $\frac{\Delta \mathrm{v}}{\mathrm{v}_{0}} \times 100=\frac{0.414 \sqrt{\mathrm{gr}}}{\sqrt{\mathrm{gr}}} \times 100$ $\Delta \mathrm{v}=41.4 \%$
BCECE-2003
Gravitation
138739
A body of mass $m$ is situated on the earth in the gravitational field of sun. For the body to escape from the gravitation pull of the solar system the body must be imparted an escape velocity of (assume earth to be stationary)
D Given that, Radius of a planet $=\mathrm{R}$ Density of planet $=\rho$ Escape velocity on its surface $\left(\mathrm{v}_{\mathrm{e}}\right)=$ ? We know that, Escape velocity \(\left(v_e\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \quad\left(\begin{array}{c}\mathrm{M}=\mathrm{V} \times \rho \\ =\frac{4}{3} \pi \mathrm{R}^3 \times \rho\end{array}\right)\) $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\mathrm{R} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho}$
BCECE-2011
Gravitation
138736
If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth, the height of the satellite above the surface of the earth is-
1 $2 \mathrm{R}$
2 $\frac{\mathrm{R}}{2}$
3 $\mathrm{R}$
4 $\frac{\mathrm{R}}{4}$
Explanation:
C According to question, Orbital velocity, $\mathrm{v}_{\mathrm{o}}=\frac{\mathrm{v}_{\mathrm{e}}}{2}=\frac{1}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \quad\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\right)$ $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}} \ldots . . \text { (i) }$ We know that, orbital velocity at height $\mathrm{h}$, $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}} .$ On comparing equation (i) and (ii) - $\sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}}$ Squaring both sides, $\frac{\mathrm{GM}}{2 \mathrm{R}}=\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}$ So, $\mathrm{R}+\mathrm{h}=2 \mathrm{R}$ $\mathrm{h}=2 \mathrm{R}-\mathrm{R}$ $\mathrm{h}=\mathrm{R}$
BCECE-2011
Gravitation
138737
The escape velocity of a projectile on the earth's surface is $11.2 \mathrm{kms}^{-1}$. A body is projected out with thrice this speed. The speed of the body far away from the earth will be :
1 $22.4 \mathrm{kms}^{-1}$
2 $31.7 \mathrm{kms}^{-1}$
3 $33.6 \mathrm{kms}^{-1}$
4 none of these
Explanation:
B Given that, $\mathrm{v}_{\mathrm{e}}=11.2 \mathrm{kms}^{-1}$ $\mathrm{v}_{\mathrm{P}}=3 \times \mathrm{v}_{\mathrm{e}}$ Mass of body $=\mathrm{m}$ Let velocity of the projectile far away of the earth $=v_{f}$ $\text { Total energy }=\frac{1}{2} \mathrm{mv}_{\mathrm{P}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Now, From the law of conservation of energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{\left(\mathrm{v}_{\mathrm{p}}\right)^{2}-\left(\mathrm{v}_{\mathrm{e}}\right)^{2}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{(3 \times 11.2)^{2}-(11.2)^{2}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{8} \times 11.2$ $\mathrm{v}_{\mathrm{f}}=31.68 \mathrm{kms}^{-1}$
BHU 2006
Gravitation
138738
A satellite is orbiting around the earth. By what percentage should we increase its velocity, so as to enable it escape away from the earth ?
1 $41.4 \%$
2 $50 \%$
3 $82.8 \%$
4 $100 \%$
Explanation:
A We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gr}}$ Orbital velocity $\left(\mathrm{v}_{\mathrm{o}}\right)=\sqrt{\mathrm{gr}}$ So, the increase in velocity $\Delta \mathrm{v} =\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{\mathrm{o}}$ $=\sqrt{2 \mathrm{gr}}-\sqrt{\mathrm{gr}}$ $\Delta \mathrm{v} =(\sqrt{2}-1) \sqrt{\mathrm{gr}}$ $\Delta \mathrm{v} =0.414 \sqrt{\mathrm{gr}}$ Now, $\frac{\Delta \mathrm{v}}{\mathrm{v}_{0}} \times 100=\frac{\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}_{\mathrm{o}}} \times 100$ $\frac{\Delta \mathrm{v}}{\mathrm{v}_{0}} \times 100=\frac{0.414 \sqrt{\mathrm{gr}}}{\sqrt{\mathrm{gr}}} \times 100$ $\Delta \mathrm{v}=41.4 \%$
BCECE-2003
Gravitation
138739
A body of mass $m$ is situated on the earth in the gravitational field of sun. For the body to escape from the gravitation pull of the solar system the body must be imparted an escape velocity of (assume earth to be stationary)
