138691
If the radius of the earth shrinks by $0.2 \%$ without any change in the mass. The escape velocity from the surface of earth is
1 decreased by $0.1 \%$
2 decreased by $0.4 \%$
3 increased by $0.1 \%$
4 increased by $0.4 \%$
Explanation:
C We know that, Escape velocity $8 \sqrt{2}$ Where, $\quad \mathrm{M}=$ moss of earth $\quad \mathrm{R}=$ Radius of earth $\ln \mathrm{v}_{\mathrm{e}}=\ln \left(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\right)$ $\ln \mathrm{v}_{\mathrm{e}}=\ln \left(\frac{2 \mathrm{GM}}{\mathrm{R}}\right)^{1 / 2}$ $\ln \mathrm{v}_{\mathrm{e}}=\frac{1}{2} \ln \frac{2 \mathrm{GM}}{\mathrm{R}}$ $\ln \mathrm{v}_{\mathrm{e}}=\frac{1}{2}[\ln 2+\ln \mathrm{G}+\ln \mathrm{M}-\ln \mathrm{R}][\ln \mathrm{a} \times \mathrm{b}=\ln \mathrm{a}+\ln \mathrm{b}]$ Differentiate both side: $\frac{1}{\mathrm{v}_{\mathrm{e}}} \mathrm{dv}_{\mathrm{e}}=\frac{1}{2}\left[0+0-\frac{1}{\mathrm{R}} \mathrm{dR}\right]$ $\frac{\mathrm{dv}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}}=-\frac{1}{2} \frac{\mathrm{dR}}{\mathrm{R}}$ $\frac{\Delta \mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}}=-\frac{1}{2} \frac{\Delta \mathrm{R}}{\mathrm{R}}$ $\frac{\Delta \mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}} \times 100=-\frac{1}{2} \frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100$ $\left(\%\right.$ change in $\left.\mathrm{v}_{\mathrm{e}}\right)=-\frac{1}{2}(\%$ change $)$ $=-\frac{1}{2} \times[-0.2] \quad\left[\therefore \frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100=-0.2\right]$ $=0.1$ So, escape velocity from the surface of earth is increased by $0.1 \%$.
CG PET 2019
Gravitation
138693
The escape velocity of a particle of mass $m$ varies as
1 $\mathrm{m}^{2}$
2 $\mathrm{m}$
3 $\mathrm{m}^{0}$
4 $\mathrm{m}^{-1}$
Explanation:
C We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{G}=$ universal gravitational constant $M=$ Mass of earth $\mathrm{R}=$ Radius Hence, it is clear that escape velocity is independent of mass (m) of particle.
UPCPMT 1978
Gravitation
138694
Rohini satellite at a height of $500 \mathrm{~km}$ and INSAT-B at a height of $3600 \mathrm{~km}$ from surface of earth, then relation between their orbital velocity $\left(v_{R}, v_{I}\right)$ is
A We know that, Orbital velocity of a satellite is given by. $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}}}$ Where, $\mathrm{G}=$ gravitational constant $\mathrm{M}_{\mathrm{e}}=$ mass of earth $\mathrm{R}=$ distance of the satellite from the center of earth $\therefore \quad \mathrm{v}_{\mathrm{o}} \propto \frac{1}{\sqrt{\mathrm{R}}}$ As, the distance of INSAT - B from the centre and the earth is greater than the distance of Rohini from the centre of the earth. $\therefore \quad \mathrm{v}_{\mathrm{R}}>\mathrm{v}_{\mathrm{I}}$
CG PET- 2007
Gravitation
138695
If the escape velocity for a monkey from the earth surface is $11.2 \mathrm{~km} / \mathrm{s}$, then the escape velocity for the elephant is
1 less than $11.2 \mathrm{Km} / \mathrm{s}$
2 more than $11.2 \mathrm{Km} / \mathrm{s}$
3 $11.2 \mathrm{Km} / \mathrm{s}$
4 None of the above
Explanation:
C According to question, the minimum velocity required to escape the gravitational field of the earth is called escape velocity. We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ Escape velocity is not depend upon mass of projected body. Hence, escape velocity of monkey and elephant will be same $(11.2 \mathrm{~km} / \mathrm{s})$.
138691
If the radius of the earth shrinks by $0.2 \%$ without any change in the mass. The escape velocity from the surface of earth is
1 decreased by $0.1 \%$
2 decreased by $0.4 \%$
3 increased by $0.1 \%$
4 increased by $0.4 \%$
Explanation:
C We know that, Escape velocity $8 \sqrt{2}$ Where, $\quad \mathrm{M}=$ moss of earth $\quad \mathrm{R}=$ Radius of earth $\ln \mathrm{v}_{\mathrm{e}}=\ln \left(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\right)$ $\ln \mathrm{v}_{\mathrm{e}}=\ln \left(\frac{2 \mathrm{GM}}{\mathrm{R}}\right)^{1 / 2}$ $\ln \mathrm{v}_{\mathrm{e}}=\frac{1}{2} \ln \frac{2 \mathrm{GM}}{\mathrm{R}}$ $\ln \mathrm{v}_{\mathrm{e}}=\frac{1}{2}[\ln 2+\ln \mathrm{G}+\ln \mathrm{M}-\ln \mathrm{R}][\ln \mathrm{a} \times \mathrm{b}=\ln \mathrm{a}+\ln \mathrm{b}]$ Differentiate both side: $\frac{1}{\mathrm{v}_{\mathrm{e}}} \mathrm{dv}_{\mathrm{e}}=\frac{1}{2}\left[0+0-\frac{1}{\mathrm{R}} \mathrm{dR}\right]$ $\frac{\mathrm{dv}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}}=-\frac{1}{2} \frac{\mathrm{dR}}{\mathrm{R}}$ $\frac{\Delta \mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}}=-\frac{1}{2} \frac{\Delta \mathrm{R}}{\mathrm{R}}$ $\frac{\Delta \mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}} \times 100=-\frac{1}{2} \frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100$ $\left(\%\right.$ change in $\left.\mathrm{v}_{\mathrm{e}}\right)=-\frac{1}{2}(\%$ change $)$ $=-\frac{1}{2} \times[-0.2] \quad\left[\therefore \frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100=-0.2\right]$ $=0.1$ So, escape velocity from the surface of earth is increased by $0.1 \%$.
CG PET 2019
Gravitation
138693
The escape velocity of a particle of mass $m$ varies as
1 $\mathrm{m}^{2}$
2 $\mathrm{m}$
3 $\mathrm{m}^{0}$
4 $\mathrm{m}^{-1}$
Explanation:
C We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{G}=$ universal gravitational constant $M=$ Mass of earth $\mathrm{R}=$ Radius Hence, it is clear that escape velocity is independent of mass (m) of particle.
UPCPMT 1978
Gravitation
138694
Rohini satellite at a height of $500 \mathrm{~km}$ and INSAT-B at a height of $3600 \mathrm{~km}$ from surface of earth, then relation between their orbital velocity $\left(v_{R}, v_{I}\right)$ is
A We know that, Orbital velocity of a satellite is given by. $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}}}$ Where, $\mathrm{G}=$ gravitational constant $\mathrm{M}_{\mathrm{e}}=$ mass of earth $\mathrm{R}=$ distance of the satellite from the center of earth $\therefore \quad \mathrm{v}_{\mathrm{o}} \propto \frac{1}{\sqrt{\mathrm{R}}}$ As, the distance of INSAT - B from the centre and the earth is greater than the distance of Rohini from the centre of the earth. $\therefore \quad \mathrm{v}_{\mathrm{R}}>\mathrm{v}_{\mathrm{I}}$
CG PET- 2007
Gravitation
138695
If the escape velocity for a monkey from the earth surface is $11.2 \mathrm{~km} / \mathrm{s}$, then the escape velocity for the elephant is
1 less than $11.2 \mathrm{Km} / \mathrm{s}$
2 more than $11.2 \mathrm{Km} / \mathrm{s}$
3 $11.2 \mathrm{Km} / \mathrm{s}$
4 None of the above
Explanation:
C According to question, the minimum velocity required to escape the gravitational field of the earth is called escape velocity. We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ Escape velocity is not depend upon mass of projected body. Hence, escape velocity of monkey and elephant will be same $(11.2 \mathrm{~km} / \mathrm{s})$.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138691
If the radius of the earth shrinks by $0.2 \%$ without any change in the mass. The escape velocity from the surface of earth is
1 decreased by $0.1 \%$
2 decreased by $0.4 \%$
3 increased by $0.1 \%$
4 increased by $0.4 \%$
Explanation:
C We know that, Escape velocity $8 \sqrt{2}$ Where, $\quad \mathrm{M}=$ moss of earth $\quad \mathrm{R}=$ Radius of earth $\ln \mathrm{v}_{\mathrm{e}}=\ln \left(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\right)$ $\ln \mathrm{v}_{\mathrm{e}}=\ln \left(\frac{2 \mathrm{GM}}{\mathrm{R}}\right)^{1 / 2}$ $\ln \mathrm{v}_{\mathrm{e}}=\frac{1}{2} \ln \frac{2 \mathrm{GM}}{\mathrm{R}}$ $\ln \mathrm{v}_{\mathrm{e}}=\frac{1}{2}[\ln 2+\ln \mathrm{G}+\ln \mathrm{M}-\ln \mathrm{R}][\ln \mathrm{a} \times \mathrm{b}=\ln \mathrm{a}+\ln \mathrm{b}]$ Differentiate both side: $\frac{1}{\mathrm{v}_{\mathrm{e}}} \mathrm{dv}_{\mathrm{e}}=\frac{1}{2}\left[0+0-\frac{1}{\mathrm{R}} \mathrm{dR}\right]$ $\frac{\mathrm{dv}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}}=-\frac{1}{2} \frac{\mathrm{dR}}{\mathrm{R}}$ $\frac{\Delta \mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}}=-\frac{1}{2} \frac{\Delta \mathrm{R}}{\mathrm{R}}$ $\frac{\Delta \mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}} \times 100=-\frac{1}{2} \frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100$ $\left(\%\right.$ change in $\left.\mathrm{v}_{\mathrm{e}}\right)=-\frac{1}{2}(\%$ change $)$ $=-\frac{1}{2} \times[-0.2] \quad\left[\therefore \frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100=-0.2\right]$ $=0.1$ So, escape velocity from the surface of earth is increased by $0.1 \%$.
CG PET 2019
Gravitation
138693
The escape velocity of a particle of mass $m$ varies as
1 $\mathrm{m}^{2}$
2 $\mathrm{m}$
3 $\mathrm{m}^{0}$
4 $\mathrm{m}^{-1}$
Explanation:
C We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{G}=$ universal gravitational constant $M=$ Mass of earth $\mathrm{R}=$ Radius Hence, it is clear that escape velocity is independent of mass (m) of particle.
UPCPMT 1978
Gravitation
138694
Rohini satellite at a height of $500 \mathrm{~km}$ and INSAT-B at a height of $3600 \mathrm{~km}$ from surface of earth, then relation between their orbital velocity $\left(v_{R}, v_{I}\right)$ is
A We know that, Orbital velocity of a satellite is given by. $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}}}$ Where, $\mathrm{G}=$ gravitational constant $\mathrm{M}_{\mathrm{e}}=$ mass of earth $\mathrm{R}=$ distance of the satellite from the center of earth $\therefore \quad \mathrm{v}_{\mathrm{o}} \propto \frac{1}{\sqrt{\mathrm{R}}}$ As, the distance of INSAT - B from the centre and the earth is greater than the distance of Rohini from the centre of the earth. $\therefore \quad \mathrm{v}_{\mathrm{R}}>\mathrm{v}_{\mathrm{I}}$
CG PET- 2007
Gravitation
138695
If the escape velocity for a monkey from the earth surface is $11.2 \mathrm{~km} / \mathrm{s}$, then the escape velocity for the elephant is
1 less than $11.2 \mathrm{Km} / \mathrm{s}$
2 more than $11.2 \mathrm{Km} / \mathrm{s}$
3 $11.2 \mathrm{Km} / \mathrm{s}$
4 None of the above
Explanation:
C According to question, the minimum velocity required to escape the gravitational field of the earth is called escape velocity. We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ Escape velocity is not depend upon mass of projected body. Hence, escape velocity of monkey and elephant will be same $(11.2 \mathrm{~km} / \mathrm{s})$.
138691
If the radius of the earth shrinks by $0.2 \%$ without any change in the mass. The escape velocity from the surface of earth is
1 decreased by $0.1 \%$
2 decreased by $0.4 \%$
3 increased by $0.1 \%$
4 increased by $0.4 \%$
Explanation:
C We know that, Escape velocity $8 \sqrt{2}$ Where, $\quad \mathrm{M}=$ moss of earth $\quad \mathrm{R}=$ Radius of earth $\ln \mathrm{v}_{\mathrm{e}}=\ln \left(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\right)$ $\ln \mathrm{v}_{\mathrm{e}}=\ln \left(\frac{2 \mathrm{GM}}{\mathrm{R}}\right)^{1 / 2}$ $\ln \mathrm{v}_{\mathrm{e}}=\frac{1}{2} \ln \frac{2 \mathrm{GM}}{\mathrm{R}}$ $\ln \mathrm{v}_{\mathrm{e}}=\frac{1}{2}[\ln 2+\ln \mathrm{G}+\ln \mathrm{M}-\ln \mathrm{R}][\ln \mathrm{a} \times \mathrm{b}=\ln \mathrm{a}+\ln \mathrm{b}]$ Differentiate both side: $\frac{1}{\mathrm{v}_{\mathrm{e}}} \mathrm{dv}_{\mathrm{e}}=\frac{1}{2}\left[0+0-\frac{1}{\mathrm{R}} \mathrm{dR}\right]$ $\frac{\mathrm{dv}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}}=-\frac{1}{2} \frac{\mathrm{dR}}{\mathrm{R}}$ $\frac{\Delta \mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}}=-\frac{1}{2} \frac{\Delta \mathrm{R}}{\mathrm{R}}$ $\frac{\Delta \mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}} \times 100=-\frac{1}{2} \frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100$ $\left(\%\right.$ change in $\left.\mathrm{v}_{\mathrm{e}}\right)=-\frac{1}{2}(\%$ change $)$ $=-\frac{1}{2} \times[-0.2] \quad\left[\therefore \frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100=-0.2\right]$ $=0.1$ So, escape velocity from the surface of earth is increased by $0.1 \%$.
CG PET 2019
Gravitation
138693
The escape velocity of a particle of mass $m$ varies as
1 $\mathrm{m}^{2}$
2 $\mathrm{m}$
3 $\mathrm{m}^{0}$
4 $\mathrm{m}^{-1}$
Explanation:
C We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{G}=$ universal gravitational constant $M=$ Mass of earth $\mathrm{R}=$ Radius Hence, it is clear that escape velocity is independent of mass (m) of particle.
UPCPMT 1978
Gravitation
138694
Rohini satellite at a height of $500 \mathrm{~km}$ and INSAT-B at a height of $3600 \mathrm{~km}$ from surface of earth, then relation between their orbital velocity $\left(v_{R}, v_{I}\right)$ is
A We know that, Orbital velocity of a satellite is given by. $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}}}$ Where, $\mathrm{G}=$ gravitational constant $\mathrm{M}_{\mathrm{e}}=$ mass of earth $\mathrm{R}=$ distance of the satellite from the center of earth $\therefore \quad \mathrm{v}_{\mathrm{o}} \propto \frac{1}{\sqrt{\mathrm{R}}}$ As, the distance of INSAT - B from the centre and the earth is greater than the distance of Rohini from the centre of the earth. $\therefore \quad \mathrm{v}_{\mathrm{R}}>\mathrm{v}_{\mathrm{I}}$
CG PET- 2007
Gravitation
138695
If the escape velocity for a monkey from the earth surface is $11.2 \mathrm{~km} / \mathrm{s}$, then the escape velocity for the elephant is
1 less than $11.2 \mathrm{Km} / \mathrm{s}$
2 more than $11.2 \mathrm{Km} / \mathrm{s}$
3 $11.2 \mathrm{Km} / \mathrm{s}$
4 None of the above
Explanation:
C According to question, the minimum velocity required to escape the gravitational field of the earth is called escape velocity. We know that, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ Escape velocity is not depend upon mass of projected body. Hence, escape velocity of monkey and elephant will be same $(11.2 \mathrm{~km} / \mathrm{s})$.