138672
If the escape velocity of a planet is 3 times that of the earth and its radius is 4 times that of the earth, then the mass of the planet is ( mass of the earth $=6 \times 10^{24} \mathrm{~kg}$ )
1 $1.62 \times 10^{22} \mathrm{~kg}$
2 $0.72 \times 10^{22} \mathrm{~kg}$
3 $2.16 \times 10^{26} \mathrm{~kg}$
4 $1.22 \times 10^{22} \mathrm{~kg}$
5 $3.6 \times 10^{22} \mathrm{~kg}$
Explanation:
C We know, escape velocity of the planet is $\mathrm{v}_{\mathrm{p}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}}$ Escape velocity of the earth is $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ According to given question, $\mathrm{v}_{\mathrm{p}}=3 \mathrm{v}_{\mathrm{e}} \text { and } \mathrm{R}_{\mathrm{p}}=4 \mathrm{R}_{\mathrm{e}}$ From equation (ii) $\because \quad \mathrm{v}_{\mathrm{p}}=3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From equation (i) and (iii) $\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}} =3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{4 \mathrm{R}_{\mathrm{e}}}} =3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\frac{1}{2} \sqrt{\mathrm{M}_{\mathrm{p}}} =3 \sqrt{\mathrm{M}_{\mathrm{e}}}$ $\frac{1}{4} \mathrm{M}_{\mathrm{p}} =9 \mathrm{M}_{\mathrm{e}}$ $\mathrm{M}_{\mathrm{p}}=36 \mathrm{M}_{\mathrm{e}} =36 \times 6 \times 10^{24} \mathrm{~kg}$ $\mathrm{M}_{\mathrm{p}} =216 \times 10^{24} \mathrm{~kg}$ $\mathrm{M}_{\mathrm{p}} =2.16 \times 10^{26} \mathrm{~kg}$
MP PET 1998
Gravitation
138673
A satellite is revolving around the earth with a kinetic energy $E$. The minimum addition of kinetic energy needed to make it escape from its orbit is
1 $2 \mathrm{E}$
2 $\sqrt{\mathrm{E}}$
3 $\mathrm{E} / 2$
4 $\sqrt{\mathrm{E}} / 2$
5 $\mathrm{E}$
Explanation:
E Orbital velocity of the satellite $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$ Kinetic energy of satellite in its orbit, $\mathrm{E}=\frac{1}{2} \mathrm{mv}_{\mathrm{o}}^{2}=\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}$ Kinetic energy for escaping, $\mathrm{E}^{\prime}=\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ $=\frac{1}{2} \times \frac{2 \mathrm{GMm}}{\mathrm{r}}=\frac{\mathrm{GMm}}{\mathrm{r}} \quad\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{r}}}\right)$ $\mathrm{E}^{\prime}=2 \times \frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}$ $\mathrm{E}^{\prime}=2 \mathrm{E}$ $\therefore$ Additional kinetic energy required $=2 \mathrm{E}-\mathrm{E}$ $=\mathrm{E}$
Kerala CEE - 2008
Gravitation
138674
Three identical bodies of mass $M$ are located at the vertices of an equilateral triangle of side $L$. they revolve under the effect of mutual gravitational force in a circular orbit, circumscribing the triangle while preserving the equilateral triangle. Their orbital velocity is
1 $\sqrt{\frac{\mathrm{GM}}{\mathrm{L}}}$
2 $\sqrt{\frac{3 \mathrm{GM}}{2 \mathrm{~L}}}$
3 $\sqrt{\frac{3 \mathrm{GM}}{\mathrm{L}}}$
4 $\sqrt{\frac{2 \mathrm{GM}}{3 \mathrm{~L}}}$
5 $\sqrt{\frac{\mathrm{GM}}{3 \mathrm{~L}}}$
Explanation:
A According to figure, $\frac{L}{2}=r \cos 30^{\circ}=\frac{\sqrt{3}}{2} r$ $r=\frac{L}{\sqrt{3}}$ The net force of attraction on mass at A due to masses at $\mathrm{B}$ and $\mathrm{C}$ is $\mathrm{F}_{\mathrm{A}} =2\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}}\right] \cos 30^{\circ} \quad\left\{\because \mathrm{F}=\frac{\mathrm{Gm}_{1} \cdot \mathrm{m}_{2}}{\mathrm{~d}^{2}}\right\}$ $=2\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \cdot \frac{\sqrt{3}}{2}\right]=\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \sqrt{3}\right]$ Centripetal force for circumscribing the triangle in a circular orbit is provided by mutual gravitational interaction. $\text { i.e. } \frac{\mathrm{MV}^{2}}{\mathrm{r}}=\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \sqrt{3}$ $\text { Or } \frac{\mathrm{MV}^{2} \sqrt{3}}{\mathrm{~L}}=\frac{\mathrm{GM}^{2} \sqrt{3}}{\mathrm{~L}^{2}}$ $\therefore \mathrm{V}=\sqrt{\frac{\mathrm{GM}}{\mathrm{L}}}$
Kerala CEE - 2008
Gravitation
138675
The escape velocity of body on the surface of earth is $11.2 \mathrm{~km} / \mathrm{s}$. If the mass of the earth is doubled and its radius halved, the escape velocity becomes
1 $5.6 \mathrm{~km} / \mathrm{s}$
2 $11.2 \mathrm{~km} / \mathrm{s}$
3 $22.4 \mathrm{~km} / \mathrm{s}$
4 $44.8 \mathrm{~km} / \mathrm{s}$
5 $67.2 \mathrm{~km} / \mathrm{s}$
Explanation:
C Given, Escape velocity of a body $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{~km} / \mathrm{s}$ New mass of the earth $\mathrm{M}_{\mathrm{e}}{ }^{\prime}=2 \mathrm{M}_{\mathrm{e}}$ New radius of the earth $\mathrm{R}_{\mathrm{e}}{ }^{\prime}=0.5 \mathrm{R}_{\mathrm{e}}$ Escape velocity $\left(v_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{e}} \propto \sqrt{\frac{\mathrm{M}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Therefore, $\frac{v_{e}}{v_{e}{ }^{\prime}}=\sqrt{\frac{M_{e}}{R_{e}} \times \frac{R_{e}^{\prime}}{M_{e}^{\prime}}}=\sqrt{\frac{M_{e}}{R_{e}} \times \frac{0.5 R_{e}}{2 M_{e}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}{ }^{\prime}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$ or, $\mathrm{v}_{\mathrm{e}}{ }^{\prime}=2 \mathrm{v}_{\mathrm{e}}=2 \times 11.2=22.4 \mathrm{~km} / \mathrm{sec}$.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138672
If the escape velocity of a planet is 3 times that of the earth and its radius is 4 times that of the earth, then the mass of the planet is ( mass of the earth $=6 \times 10^{24} \mathrm{~kg}$ )
1 $1.62 \times 10^{22} \mathrm{~kg}$
2 $0.72 \times 10^{22} \mathrm{~kg}$
3 $2.16 \times 10^{26} \mathrm{~kg}$
4 $1.22 \times 10^{22} \mathrm{~kg}$
5 $3.6 \times 10^{22} \mathrm{~kg}$
Explanation:
C We know, escape velocity of the planet is $\mathrm{v}_{\mathrm{p}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}}$ Escape velocity of the earth is $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ According to given question, $\mathrm{v}_{\mathrm{p}}=3 \mathrm{v}_{\mathrm{e}} \text { and } \mathrm{R}_{\mathrm{p}}=4 \mathrm{R}_{\mathrm{e}}$ From equation (ii) $\because \quad \mathrm{v}_{\mathrm{p}}=3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From equation (i) and (iii) $\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}} =3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{4 \mathrm{R}_{\mathrm{e}}}} =3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\frac{1}{2} \sqrt{\mathrm{M}_{\mathrm{p}}} =3 \sqrt{\mathrm{M}_{\mathrm{e}}}$ $\frac{1}{4} \mathrm{M}_{\mathrm{p}} =9 \mathrm{M}_{\mathrm{e}}$ $\mathrm{M}_{\mathrm{p}}=36 \mathrm{M}_{\mathrm{e}} =36 \times 6 \times 10^{24} \mathrm{~kg}$ $\mathrm{M}_{\mathrm{p}} =216 \times 10^{24} \mathrm{~kg}$ $\mathrm{M}_{\mathrm{p}} =2.16 \times 10^{26} \mathrm{~kg}$
MP PET 1998
Gravitation
138673
A satellite is revolving around the earth with a kinetic energy $E$. The minimum addition of kinetic energy needed to make it escape from its orbit is
1 $2 \mathrm{E}$
2 $\sqrt{\mathrm{E}}$
3 $\mathrm{E} / 2$
4 $\sqrt{\mathrm{E}} / 2$
5 $\mathrm{E}$
Explanation:
E Orbital velocity of the satellite $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$ Kinetic energy of satellite in its orbit, $\mathrm{E}=\frac{1}{2} \mathrm{mv}_{\mathrm{o}}^{2}=\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}$ Kinetic energy for escaping, $\mathrm{E}^{\prime}=\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ $=\frac{1}{2} \times \frac{2 \mathrm{GMm}}{\mathrm{r}}=\frac{\mathrm{GMm}}{\mathrm{r}} \quad\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{r}}}\right)$ $\mathrm{E}^{\prime}=2 \times \frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}$ $\mathrm{E}^{\prime}=2 \mathrm{E}$ $\therefore$ Additional kinetic energy required $=2 \mathrm{E}-\mathrm{E}$ $=\mathrm{E}$
Kerala CEE - 2008
Gravitation
138674
Three identical bodies of mass $M$ are located at the vertices of an equilateral triangle of side $L$. they revolve under the effect of mutual gravitational force in a circular orbit, circumscribing the triangle while preserving the equilateral triangle. Their orbital velocity is
1 $\sqrt{\frac{\mathrm{GM}}{\mathrm{L}}}$
2 $\sqrt{\frac{3 \mathrm{GM}}{2 \mathrm{~L}}}$
3 $\sqrt{\frac{3 \mathrm{GM}}{\mathrm{L}}}$
4 $\sqrt{\frac{2 \mathrm{GM}}{3 \mathrm{~L}}}$
5 $\sqrt{\frac{\mathrm{GM}}{3 \mathrm{~L}}}$
Explanation:
A According to figure, $\frac{L}{2}=r \cos 30^{\circ}=\frac{\sqrt{3}}{2} r$ $r=\frac{L}{\sqrt{3}}$ The net force of attraction on mass at A due to masses at $\mathrm{B}$ and $\mathrm{C}$ is $\mathrm{F}_{\mathrm{A}} =2\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}}\right] \cos 30^{\circ} \quad\left\{\because \mathrm{F}=\frac{\mathrm{Gm}_{1} \cdot \mathrm{m}_{2}}{\mathrm{~d}^{2}}\right\}$ $=2\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \cdot \frac{\sqrt{3}}{2}\right]=\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \sqrt{3}\right]$ Centripetal force for circumscribing the triangle in a circular orbit is provided by mutual gravitational interaction. $\text { i.e. } \frac{\mathrm{MV}^{2}}{\mathrm{r}}=\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \sqrt{3}$ $\text { Or } \frac{\mathrm{MV}^{2} \sqrt{3}}{\mathrm{~L}}=\frac{\mathrm{GM}^{2} \sqrt{3}}{\mathrm{~L}^{2}}$ $\therefore \mathrm{V}=\sqrt{\frac{\mathrm{GM}}{\mathrm{L}}}$
Kerala CEE - 2008
Gravitation
138675
The escape velocity of body on the surface of earth is $11.2 \mathrm{~km} / \mathrm{s}$. If the mass of the earth is doubled and its radius halved, the escape velocity becomes
1 $5.6 \mathrm{~km} / \mathrm{s}$
2 $11.2 \mathrm{~km} / \mathrm{s}$
3 $22.4 \mathrm{~km} / \mathrm{s}$
4 $44.8 \mathrm{~km} / \mathrm{s}$
5 $67.2 \mathrm{~km} / \mathrm{s}$
Explanation:
C Given, Escape velocity of a body $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{~km} / \mathrm{s}$ New mass of the earth $\mathrm{M}_{\mathrm{e}}{ }^{\prime}=2 \mathrm{M}_{\mathrm{e}}$ New radius of the earth $\mathrm{R}_{\mathrm{e}}{ }^{\prime}=0.5 \mathrm{R}_{\mathrm{e}}$ Escape velocity $\left(v_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{e}} \propto \sqrt{\frac{\mathrm{M}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Therefore, $\frac{v_{e}}{v_{e}{ }^{\prime}}=\sqrt{\frac{M_{e}}{R_{e}} \times \frac{R_{e}^{\prime}}{M_{e}^{\prime}}}=\sqrt{\frac{M_{e}}{R_{e}} \times \frac{0.5 R_{e}}{2 M_{e}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}{ }^{\prime}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$ or, $\mathrm{v}_{\mathrm{e}}{ }^{\prime}=2 \mathrm{v}_{\mathrm{e}}=2 \times 11.2=22.4 \mathrm{~km} / \mathrm{sec}$.
138672
If the escape velocity of a planet is 3 times that of the earth and its radius is 4 times that of the earth, then the mass of the planet is ( mass of the earth $=6 \times 10^{24} \mathrm{~kg}$ )
1 $1.62 \times 10^{22} \mathrm{~kg}$
2 $0.72 \times 10^{22} \mathrm{~kg}$
3 $2.16 \times 10^{26} \mathrm{~kg}$
4 $1.22 \times 10^{22} \mathrm{~kg}$
5 $3.6 \times 10^{22} \mathrm{~kg}$
Explanation:
C We know, escape velocity of the planet is $\mathrm{v}_{\mathrm{p}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}}$ Escape velocity of the earth is $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ According to given question, $\mathrm{v}_{\mathrm{p}}=3 \mathrm{v}_{\mathrm{e}} \text { and } \mathrm{R}_{\mathrm{p}}=4 \mathrm{R}_{\mathrm{e}}$ From equation (ii) $\because \quad \mathrm{v}_{\mathrm{p}}=3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From equation (i) and (iii) $\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}} =3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{4 \mathrm{R}_{\mathrm{e}}}} =3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\frac{1}{2} \sqrt{\mathrm{M}_{\mathrm{p}}} =3 \sqrt{\mathrm{M}_{\mathrm{e}}}$ $\frac{1}{4} \mathrm{M}_{\mathrm{p}} =9 \mathrm{M}_{\mathrm{e}}$ $\mathrm{M}_{\mathrm{p}}=36 \mathrm{M}_{\mathrm{e}} =36 \times 6 \times 10^{24} \mathrm{~kg}$ $\mathrm{M}_{\mathrm{p}} =216 \times 10^{24} \mathrm{~kg}$ $\mathrm{M}_{\mathrm{p}} =2.16 \times 10^{26} \mathrm{~kg}$
MP PET 1998
Gravitation
138673
A satellite is revolving around the earth with a kinetic energy $E$. The minimum addition of kinetic energy needed to make it escape from its orbit is
1 $2 \mathrm{E}$
2 $\sqrt{\mathrm{E}}$
3 $\mathrm{E} / 2$
4 $\sqrt{\mathrm{E}} / 2$
5 $\mathrm{E}$
Explanation:
E Orbital velocity of the satellite $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$ Kinetic energy of satellite in its orbit, $\mathrm{E}=\frac{1}{2} \mathrm{mv}_{\mathrm{o}}^{2}=\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}$ Kinetic energy for escaping, $\mathrm{E}^{\prime}=\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ $=\frac{1}{2} \times \frac{2 \mathrm{GMm}}{\mathrm{r}}=\frac{\mathrm{GMm}}{\mathrm{r}} \quad\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{r}}}\right)$ $\mathrm{E}^{\prime}=2 \times \frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}$ $\mathrm{E}^{\prime}=2 \mathrm{E}$ $\therefore$ Additional kinetic energy required $=2 \mathrm{E}-\mathrm{E}$ $=\mathrm{E}$
Kerala CEE - 2008
Gravitation
138674
Three identical bodies of mass $M$ are located at the vertices of an equilateral triangle of side $L$. they revolve under the effect of mutual gravitational force in a circular orbit, circumscribing the triangle while preserving the equilateral triangle. Their orbital velocity is
1 $\sqrt{\frac{\mathrm{GM}}{\mathrm{L}}}$
2 $\sqrt{\frac{3 \mathrm{GM}}{2 \mathrm{~L}}}$
3 $\sqrt{\frac{3 \mathrm{GM}}{\mathrm{L}}}$
4 $\sqrt{\frac{2 \mathrm{GM}}{3 \mathrm{~L}}}$
5 $\sqrt{\frac{\mathrm{GM}}{3 \mathrm{~L}}}$
Explanation:
A According to figure, $\frac{L}{2}=r \cos 30^{\circ}=\frac{\sqrt{3}}{2} r$ $r=\frac{L}{\sqrt{3}}$ The net force of attraction on mass at A due to masses at $\mathrm{B}$ and $\mathrm{C}$ is $\mathrm{F}_{\mathrm{A}} =2\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}}\right] \cos 30^{\circ} \quad\left\{\because \mathrm{F}=\frac{\mathrm{Gm}_{1} \cdot \mathrm{m}_{2}}{\mathrm{~d}^{2}}\right\}$ $=2\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \cdot \frac{\sqrt{3}}{2}\right]=\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \sqrt{3}\right]$ Centripetal force for circumscribing the triangle in a circular orbit is provided by mutual gravitational interaction. $\text { i.e. } \frac{\mathrm{MV}^{2}}{\mathrm{r}}=\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \sqrt{3}$ $\text { Or } \frac{\mathrm{MV}^{2} \sqrt{3}}{\mathrm{~L}}=\frac{\mathrm{GM}^{2} \sqrt{3}}{\mathrm{~L}^{2}}$ $\therefore \mathrm{V}=\sqrt{\frac{\mathrm{GM}}{\mathrm{L}}}$
Kerala CEE - 2008
Gravitation
138675
The escape velocity of body on the surface of earth is $11.2 \mathrm{~km} / \mathrm{s}$. If the mass of the earth is doubled and its radius halved, the escape velocity becomes
1 $5.6 \mathrm{~km} / \mathrm{s}$
2 $11.2 \mathrm{~km} / \mathrm{s}$
3 $22.4 \mathrm{~km} / \mathrm{s}$
4 $44.8 \mathrm{~km} / \mathrm{s}$
5 $67.2 \mathrm{~km} / \mathrm{s}$
Explanation:
C Given, Escape velocity of a body $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{~km} / \mathrm{s}$ New mass of the earth $\mathrm{M}_{\mathrm{e}}{ }^{\prime}=2 \mathrm{M}_{\mathrm{e}}$ New radius of the earth $\mathrm{R}_{\mathrm{e}}{ }^{\prime}=0.5 \mathrm{R}_{\mathrm{e}}$ Escape velocity $\left(v_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{e}} \propto \sqrt{\frac{\mathrm{M}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Therefore, $\frac{v_{e}}{v_{e}{ }^{\prime}}=\sqrt{\frac{M_{e}}{R_{e}} \times \frac{R_{e}^{\prime}}{M_{e}^{\prime}}}=\sqrt{\frac{M_{e}}{R_{e}} \times \frac{0.5 R_{e}}{2 M_{e}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}{ }^{\prime}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$ or, $\mathrm{v}_{\mathrm{e}}{ }^{\prime}=2 \mathrm{v}_{\mathrm{e}}=2 \times 11.2=22.4 \mathrm{~km} / \mathrm{sec}$.
138672
If the escape velocity of a planet is 3 times that of the earth and its radius is 4 times that of the earth, then the mass of the planet is ( mass of the earth $=6 \times 10^{24} \mathrm{~kg}$ )
1 $1.62 \times 10^{22} \mathrm{~kg}$
2 $0.72 \times 10^{22} \mathrm{~kg}$
3 $2.16 \times 10^{26} \mathrm{~kg}$
4 $1.22 \times 10^{22} \mathrm{~kg}$
5 $3.6 \times 10^{22} \mathrm{~kg}$
Explanation:
C We know, escape velocity of the planet is $\mathrm{v}_{\mathrm{p}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}}$ Escape velocity of the earth is $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ According to given question, $\mathrm{v}_{\mathrm{p}}=3 \mathrm{v}_{\mathrm{e}} \text { and } \mathrm{R}_{\mathrm{p}}=4 \mathrm{R}_{\mathrm{e}}$ From equation (ii) $\because \quad \mathrm{v}_{\mathrm{p}}=3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From equation (i) and (iii) $\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}} =3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\sqrt{\frac{2 \mathrm{GM}_{\mathrm{p}}}{4 \mathrm{R}_{\mathrm{e}}}} =3 \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\frac{1}{2} \sqrt{\mathrm{M}_{\mathrm{p}}} =3 \sqrt{\mathrm{M}_{\mathrm{e}}}$ $\frac{1}{4} \mathrm{M}_{\mathrm{p}} =9 \mathrm{M}_{\mathrm{e}}$ $\mathrm{M}_{\mathrm{p}}=36 \mathrm{M}_{\mathrm{e}} =36 \times 6 \times 10^{24} \mathrm{~kg}$ $\mathrm{M}_{\mathrm{p}} =216 \times 10^{24} \mathrm{~kg}$ $\mathrm{M}_{\mathrm{p}} =2.16 \times 10^{26} \mathrm{~kg}$
MP PET 1998
Gravitation
138673
A satellite is revolving around the earth with a kinetic energy $E$. The minimum addition of kinetic energy needed to make it escape from its orbit is
1 $2 \mathrm{E}$
2 $\sqrt{\mathrm{E}}$
3 $\mathrm{E} / 2$
4 $\sqrt{\mathrm{E}} / 2$
5 $\mathrm{E}$
Explanation:
E Orbital velocity of the satellite $\mathrm{v}_{\mathrm{o}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$ Kinetic energy of satellite in its orbit, $\mathrm{E}=\frac{1}{2} \mathrm{mv}_{\mathrm{o}}^{2}=\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}$ Kinetic energy for escaping, $\mathrm{E}^{\prime}=\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ $=\frac{1}{2} \times \frac{2 \mathrm{GMm}}{\mathrm{r}}=\frac{\mathrm{GMm}}{\mathrm{r}} \quad\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{r}}}\right)$ $\mathrm{E}^{\prime}=2 \times \frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}$ $\mathrm{E}^{\prime}=2 \mathrm{E}$ $\therefore$ Additional kinetic energy required $=2 \mathrm{E}-\mathrm{E}$ $=\mathrm{E}$
Kerala CEE - 2008
Gravitation
138674
Three identical bodies of mass $M$ are located at the vertices of an equilateral triangle of side $L$. they revolve under the effect of mutual gravitational force in a circular orbit, circumscribing the triangle while preserving the equilateral triangle. Their orbital velocity is
1 $\sqrt{\frac{\mathrm{GM}}{\mathrm{L}}}$
2 $\sqrt{\frac{3 \mathrm{GM}}{2 \mathrm{~L}}}$
3 $\sqrt{\frac{3 \mathrm{GM}}{\mathrm{L}}}$
4 $\sqrt{\frac{2 \mathrm{GM}}{3 \mathrm{~L}}}$
5 $\sqrt{\frac{\mathrm{GM}}{3 \mathrm{~L}}}$
Explanation:
A According to figure, $\frac{L}{2}=r \cos 30^{\circ}=\frac{\sqrt{3}}{2} r$ $r=\frac{L}{\sqrt{3}}$ The net force of attraction on mass at A due to masses at $\mathrm{B}$ and $\mathrm{C}$ is $\mathrm{F}_{\mathrm{A}} =2\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}}\right] \cos 30^{\circ} \quad\left\{\because \mathrm{F}=\frac{\mathrm{Gm}_{1} \cdot \mathrm{m}_{2}}{\mathrm{~d}^{2}}\right\}$ $=2\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \cdot \frac{\sqrt{3}}{2}\right]=\left[\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \sqrt{3}\right]$ Centripetal force for circumscribing the triangle in a circular orbit is provided by mutual gravitational interaction. $\text { i.e. } \frac{\mathrm{MV}^{2}}{\mathrm{r}}=\frac{\mathrm{GM}^{2}}{\mathrm{~L}^{2}} \sqrt{3}$ $\text { Or } \frac{\mathrm{MV}^{2} \sqrt{3}}{\mathrm{~L}}=\frac{\mathrm{GM}^{2} \sqrt{3}}{\mathrm{~L}^{2}}$ $\therefore \mathrm{V}=\sqrt{\frac{\mathrm{GM}}{\mathrm{L}}}$
Kerala CEE - 2008
Gravitation
138675
The escape velocity of body on the surface of earth is $11.2 \mathrm{~km} / \mathrm{s}$. If the mass of the earth is doubled and its radius halved, the escape velocity becomes
1 $5.6 \mathrm{~km} / \mathrm{s}$
2 $11.2 \mathrm{~km} / \mathrm{s}$
3 $22.4 \mathrm{~km} / \mathrm{s}$
4 $44.8 \mathrm{~km} / \mathrm{s}$
5 $67.2 \mathrm{~km} / \mathrm{s}$
Explanation:
C Given, Escape velocity of a body $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{~km} / \mathrm{s}$ New mass of the earth $\mathrm{M}_{\mathrm{e}}{ }^{\prime}=2 \mathrm{M}_{\mathrm{e}}$ New radius of the earth $\mathrm{R}_{\mathrm{e}}{ }^{\prime}=0.5 \mathrm{R}_{\mathrm{e}}$ Escape velocity $\left(v_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{e}} \propto \sqrt{\frac{\mathrm{M}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Therefore, $\frac{v_{e}}{v_{e}{ }^{\prime}}=\sqrt{\frac{M_{e}}{R_{e}} \times \frac{R_{e}^{\prime}}{M_{e}^{\prime}}}=\sqrt{\frac{M_{e}}{R_{e}} \times \frac{0.5 R_{e}}{2 M_{e}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}}{ }^{\prime}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$ or, $\mathrm{v}_{\mathrm{e}}{ }^{\prime}=2 \mathrm{v}_{\mathrm{e}}=2 \times 11.2=22.4 \mathrm{~km} / \mathrm{sec}$.
