NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138744
The escape velocity of a body from the Earth is v. What will be the escape velocity of the same body from a planet whose radius is twice that of the Earth and mean density same as that of the Earth?
1 $8 \mathrm{v}$
2 $4 \mathrm{v}$
3 $2 \mathrm{v}$
4 $\mathrm{v}$
Explanation:
C Escape velocity from the Earth's surface $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G}\left[\rho \times \frac{4}{3} \pi \mathrm{R}^{3}\right]}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{8 \mathrm{G} \rho \pi \mathrm{R}^{2}}{3}}=\mathrm{R} \sqrt{\frac{8 \mathrm{G} \rho \pi}{3}}$ $\mathrm{v}_{\mathrm{e}} \propto \mathrm{R} \quad$ (For same density) So, $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{R}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{e}}} \quad$ (Given, $\mathrm{R}_{\mathrm{p}}=2 \mathrm{R}_{\mathrm{e}}$ ) $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{2 \mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}$ $\mathrm{v}_{\mathrm{P}}=2 \mathrm{v}_{\mathrm{e}}$ or $\quad \mathrm{v}_{\mathrm{P}}=2 \mathrm{v}$
SCRA-2010
Gravitation
138745
A spaceship is launched into a circular orbit close to the Earth's surface. What additional velocity has to be imparted to the spaceship to overcome the gravitational pull? (Radius of the Earth is R)
1 $\mathrm{gR}$
2 $\sqrt{2} \mathrm{gR}$
3 $(\sqrt{2}-1) \sqrt{\mathrm{gR}}$
4 $(\sqrt{2}-1) g R$
Explanation:
C Let the radius of Earth be $\mathrm{R}$ and acceleration due to gravity be $\mathrm{g}$. When a satellite is orbiting the earth, Orbital velocity $\left(\mathrm{v}_{0}\right)=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}=\sqrt{\mathrm{gR}} \quad\left[\therefore \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ If satellite has to overcome the gravitational pull, then its velocity should be equal to escape velocity to go in circular orbit. Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{2 \mathrm{gR}}$ Now, Additional velocity to impart to spaceship, $\mathrm{v}=\mathrm{v}_{\mathrm{e}}-\mathrm{v}_0$ $\mathrm{v}=\sqrt{2 \mathrm{gR}}-\sqrt{\mathrm{gR}}$ $\therefore \quad \mathrm{v}=(\sqrt{2}-1) \sqrt{\mathrm{gR}}$
SCRA-2010
Gravitation
138746
A satellite is orbiting close to the surface of earth. In order to make to move to infinity, its velocity must be increased by about.
1 $50 \%$
2 $40 \%$
3 $30 \%$
4 $20 \%$
Explanation:
B We know that, Orbital velocity of satellite $\left(\mathrm{v}_{\mathrm{o}}\right)=\sqrt{\mathrm{gR}}$ Escape velocity of satellite $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gR}}$ Now, $\Delta \mathrm{v} \%= \frac{\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{0}}{\mathrm{v}_{0}} \times 100$ $=\frac{(1.414-1) \sqrt{\mathrm{gR}}}{\sqrt{\mathrm{gR}}} \times 100$ $\Delta \mathrm{v} \%=41.4 \%$
SCRA-2009
Gravitation
138747
A space station is at a height equal to the radius of the Earth. If ' $v_{E}$ ' is the escape velocity on the surface of the Earth, the same on the space station is ....... time $\mathbf{v}_{\mathrm{E}}$.
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{\sqrt{2}}$
4 $\frac{1}{\sqrt{3}}$
Explanation:
C Given, We know that, $h=R_{e}$ $\text { Escape velocity }\left(\mathrm{v}_{\mathrm{E}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Now, Escape velocity at a height ' $h$ ' from the surface of Earth, $\mathrm{v}_{\mathrm{h}} =\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}+\mathrm{h}}}$ $\mathrm{v}_{\mathrm{h}} =\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}+\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{h}} =\frac{1}{\sqrt{2}} \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\therefore \quad \mathrm{v}_{\mathrm{h}} =\frac{1}{\sqrt{2}} \mathrm{v}_{\mathrm{E}}$
138744
The escape velocity of a body from the Earth is v. What will be the escape velocity of the same body from a planet whose radius is twice that of the Earth and mean density same as that of the Earth?
1 $8 \mathrm{v}$
2 $4 \mathrm{v}$
3 $2 \mathrm{v}$
4 $\mathrm{v}$
Explanation:
C Escape velocity from the Earth's surface $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G}\left[\rho \times \frac{4}{3} \pi \mathrm{R}^{3}\right]}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{8 \mathrm{G} \rho \pi \mathrm{R}^{2}}{3}}=\mathrm{R} \sqrt{\frac{8 \mathrm{G} \rho \pi}{3}}$ $\mathrm{v}_{\mathrm{e}} \propto \mathrm{R} \quad$ (For same density) So, $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{R}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{e}}} \quad$ (Given, $\mathrm{R}_{\mathrm{p}}=2 \mathrm{R}_{\mathrm{e}}$ ) $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{2 \mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}$ $\mathrm{v}_{\mathrm{P}}=2 \mathrm{v}_{\mathrm{e}}$ or $\quad \mathrm{v}_{\mathrm{P}}=2 \mathrm{v}$
SCRA-2010
Gravitation
138745
A spaceship is launched into a circular orbit close to the Earth's surface. What additional velocity has to be imparted to the spaceship to overcome the gravitational pull? (Radius of the Earth is R)
1 $\mathrm{gR}$
2 $\sqrt{2} \mathrm{gR}$
3 $(\sqrt{2}-1) \sqrt{\mathrm{gR}}$
4 $(\sqrt{2}-1) g R$
Explanation:
C Let the radius of Earth be $\mathrm{R}$ and acceleration due to gravity be $\mathrm{g}$. When a satellite is orbiting the earth, Orbital velocity $\left(\mathrm{v}_{0}\right)=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}=\sqrt{\mathrm{gR}} \quad\left[\therefore \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ If satellite has to overcome the gravitational pull, then its velocity should be equal to escape velocity to go in circular orbit. Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{2 \mathrm{gR}}$ Now, Additional velocity to impart to spaceship, $\mathrm{v}=\mathrm{v}_{\mathrm{e}}-\mathrm{v}_0$ $\mathrm{v}=\sqrt{2 \mathrm{gR}}-\sqrt{\mathrm{gR}}$ $\therefore \quad \mathrm{v}=(\sqrt{2}-1) \sqrt{\mathrm{gR}}$
SCRA-2010
Gravitation
138746
A satellite is orbiting close to the surface of earth. In order to make to move to infinity, its velocity must be increased by about.
1 $50 \%$
2 $40 \%$
3 $30 \%$
4 $20 \%$
Explanation:
B We know that, Orbital velocity of satellite $\left(\mathrm{v}_{\mathrm{o}}\right)=\sqrt{\mathrm{gR}}$ Escape velocity of satellite $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gR}}$ Now, $\Delta \mathrm{v} \%= \frac{\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{0}}{\mathrm{v}_{0}} \times 100$ $=\frac{(1.414-1) \sqrt{\mathrm{gR}}}{\sqrt{\mathrm{gR}}} \times 100$ $\Delta \mathrm{v} \%=41.4 \%$
SCRA-2009
Gravitation
138747
A space station is at a height equal to the radius of the Earth. If ' $v_{E}$ ' is the escape velocity on the surface of the Earth, the same on the space station is ....... time $\mathbf{v}_{\mathrm{E}}$.
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{\sqrt{2}}$
4 $\frac{1}{\sqrt{3}}$
Explanation:
C Given, We know that, $h=R_{e}$ $\text { Escape velocity }\left(\mathrm{v}_{\mathrm{E}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Now, Escape velocity at a height ' $h$ ' from the surface of Earth, $\mathrm{v}_{\mathrm{h}} =\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}+\mathrm{h}}}$ $\mathrm{v}_{\mathrm{h}} =\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}+\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{h}} =\frac{1}{\sqrt{2}} \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\therefore \quad \mathrm{v}_{\mathrm{h}} =\frac{1}{\sqrt{2}} \mathrm{v}_{\mathrm{E}}$
138744
The escape velocity of a body from the Earth is v. What will be the escape velocity of the same body from a planet whose radius is twice that of the Earth and mean density same as that of the Earth?
1 $8 \mathrm{v}$
2 $4 \mathrm{v}$
3 $2 \mathrm{v}$
4 $\mathrm{v}$
Explanation:
C Escape velocity from the Earth's surface $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G}\left[\rho \times \frac{4}{3} \pi \mathrm{R}^{3}\right]}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{8 \mathrm{G} \rho \pi \mathrm{R}^{2}}{3}}=\mathrm{R} \sqrt{\frac{8 \mathrm{G} \rho \pi}{3}}$ $\mathrm{v}_{\mathrm{e}} \propto \mathrm{R} \quad$ (For same density) So, $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{R}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{e}}} \quad$ (Given, $\mathrm{R}_{\mathrm{p}}=2 \mathrm{R}_{\mathrm{e}}$ ) $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{2 \mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}$ $\mathrm{v}_{\mathrm{P}}=2 \mathrm{v}_{\mathrm{e}}$ or $\quad \mathrm{v}_{\mathrm{P}}=2 \mathrm{v}$
SCRA-2010
Gravitation
138745
A spaceship is launched into a circular orbit close to the Earth's surface. What additional velocity has to be imparted to the spaceship to overcome the gravitational pull? (Radius of the Earth is R)
1 $\mathrm{gR}$
2 $\sqrt{2} \mathrm{gR}$
3 $(\sqrt{2}-1) \sqrt{\mathrm{gR}}$
4 $(\sqrt{2}-1) g R$
Explanation:
C Let the radius of Earth be $\mathrm{R}$ and acceleration due to gravity be $\mathrm{g}$. When a satellite is orbiting the earth, Orbital velocity $\left(\mathrm{v}_{0}\right)=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}=\sqrt{\mathrm{gR}} \quad\left[\therefore \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ If satellite has to overcome the gravitational pull, then its velocity should be equal to escape velocity to go in circular orbit. Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{2 \mathrm{gR}}$ Now, Additional velocity to impart to spaceship, $\mathrm{v}=\mathrm{v}_{\mathrm{e}}-\mathrm{v}_0$ $\mathrm{v}=\sqrt{2 \mathrm{gR}}-\sqrt{\mathrm{gR}}$ $\therefore \quad \mathrm{v}=(\sqrt{2}-1) \sqrt{\mathrm{gR}}$
SCRA-2010
Gravitation
138746
A satellite is orbiting close to the surface of earth. In order to make to move to infinity, its velocity must be increased by about.
1 $50 \%$
2 $40 \%$
3 $30 \%$
4 $20 \%$
Explanation:
B We know that, Orbital velocity of satellite $\left(\mathrm{v}_{\mathrm{o}}\right)=\sqrt{\mathrm{gR}}$ Escape velocity of satellite $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gR}}$ Now, $\Delta \mathrm{v} \%= \frac{\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{0}}{\mathrm{v}_{0}} \times 100$ $=\frac{(1.414-1) \sqrt{\mathrm{gR}}}{\sqrt{\mathrm{gR}}} \times 100$ $\Delta \mathrm{v} \%=41.4 \%$
SCRA-2009
Gravitation
138747
A space station is at a height equal to the radius of the Earth. If ' $v_{E}$ ' is the escape velocity on the surface of the Earth, the same on the space station is ....... time $\mathbf{v}_{\mathrm{E}}$.
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{\sqrt{2}}$
4 $\frac{1}{\sqrt{3}}$
Explanation:
C Given, We know that, $h=R_{e}$ $\text { Escape velocity }\left(\mathrm{v}_{\mathrm{E}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Now, Escape velocity at a height ' $h$ ' from the surface of Earth, $\mathrm{v}_{\mathrm{h}} =\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}+\mathrm{h}}}$ $\mathrm{v}_{\mathrm{h}} =\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}+\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{h}} =\frac{1}{\sqrt{2}} \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\therefore \quad \mathrm{v}_{\mathrm{h}} =\frac{1}{\sqrt{2}} \mathrm{v}_{\mathrm{E}}$
138744
The escape velocity of a body from the Earth is v. What will be the escape velocity of the same body from a planet whose radius is twice that of the Earth and mean density same as that of the Earth?
1 $8 \mathrm{v}$
2 $4 \mathrm{v}$
3 $2 \mathrm{v}$
4 $\mathrm{v}$
Explanation:
C Escape velocity from the Earth's surface $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G}\left[\rho \times \frac{4}{3} \pi \mathrm{R}^{3}\right]}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{8 \mathrm{G} \rho \pi \mathrm{R}^{2}}{3}}=\mathrm{R} \sqrt{\frac{8 \mathrm{G} \rho \pi}{3}}$ $\mathrm{v}_{\mathrm{e}} \propto \mathrm{R} \quad$ (For same density) So, $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{R}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{e}}} \quad$ (Given, $\mathrm{R}_{\mathrm{p}}=2 \mathrm{R}_{\mathrm{e}}$ ) $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{2 \mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}$ $\mathrm{v}_{\mathrm{P}}=2 \mathrm{v}_{\mathrm{e}}$ or $\quad \mathrm{v}_{\mathrm{P}}=2 \mathrm{v}$
SCRA-2010
Gravitation
138745
A spaceship is launched into a circular orbit close to the Earth's surface. What additional velocity has to be imparted to the spaceship to overcome the gravitational pull? (Radius of the Earth is R)
1 $\mathrm{gR}$
2 $\sqrt{2} \mathrm{gR}$
3 $(\sqrt{2}-1) \sqrt{\mathrm{gR}}$
4 $(\sqrt{2}-1) g R$
Explanation:
C Let the radius of Earth be $\mathrm{R}$ and acceleration due to gravity be $\mathrm{g}$. When a satellite is orbiting the earth, Orbital velocity $\left(\mathrm{v}_{0}\right)=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}=\sqrt{\mathrm{gR}} \quad\left[\therefore \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ If satellite has to overcome the gravitational pull, then its velocity should be equal to escape velocity to go in circular orbit. Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{2 \mathrm{gR}}$ Now, Additional velocity to impart to spaceship, $\mathrm{v}=\mathrm{v}_{\mathrm{e}}-\mathrm{v}_0$ $\mathrm{v}=\sqrt{2 \mathrm{gR}}-\sqrt{\mathrm{gR}}$ $\therefore \quad \mathrm{v}=(\sqrt{2}-1) \sqrt{\mathrm{gR}}$
SCRA-2010
Gravitation
138746
A satellite is orbiting close to the surface of earth. In order to make to move to infinity, its velocity must be increased by about.
1 $50 \%$
2 $40 \%$
3 $30 \%$
4 $20 \%$
Explanation:
B We know that, Orbital velocity of satellite $\left(\mathrm{v}_{\mathrm{o}}\right)=\sqrt{\mathrm{gR}}$ Escape velocity of satellite $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gR}}$ Now, $\Delta \mathrm{v} \%= \frac{\mathrm{v}_{\mathrm{e}}-\mathrm{v}_{0}}{\mathrm{v}_{0}} \times 100$ $=\frac{(1.414-1) \sqrt{\mathrm{gR}}}{\sqrt{\mathrm{gR}}} \times 100$ $\Delta \mathrm{v} \%=41.4 \%$
SCRA-2009
Gravitation
138747
A space station is at a height equal to the radius of the Earth. If ' $v_{E}$ ' is the escape velocity on the surface of the Earth, the same on the space station is ....... time $\mathbf{v}_{\mathrm{E}}$.
1 $\frac{1}{2}$
2 $\frac{1}{4}$
3 $\frac{1}{\sqrt{2}}$
4 $\frac{1}{\sqrt{3}}$
Explanation:
C Given, We know that, $h=R_{e}$ $\text { Escape velocity }\left(\mathrm{v}_{\mathrm{E}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Now, Escape velocity at a height ' $h$ ' from the surface of Earth, $\mathrm{v}_{\mathrm{h}} =\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}+\mathrm{h}}}$ $\mathrm{v}_{\mathrm{h}} =\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}+\mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{h}} =\frac{1}{\sqrt{2}} \sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ $\therefore \quad \mathrm{v}_{\mathrm{h}} =\frac{1}{\sqrt{2}} \mathrm{v}_{\mathrm{E}}$
