138697
The escape velocity from the earth's surface is $11.2 \mathrm{Km} / \mathrm{s}$. If a planet has a radius twice that of the earth and on which the acceleration due to gravity is twice that on the earth, then the escape velocity on this planet will be
138698
A satellite revolves very near to the earth surface. Its speed should be around
1 $5 \mathrm{~km} / \mathrm{s}$
2 $8 \mathrm{~km} / \mathrm{s}$
3 $2 \mathrm{~km} / \mathrm{s}$
4 $11 \mathrm{~km} / \mathrm{s}$
Explanation:
B Since, orbital speed of satellite near to the earth surface is approximately $8 \mathrm{~km} / \mathrm{s}$. We know that, $\text { orbital velocity }(\mathrm{v}) =\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $=\sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6400 \times 1000}}$ $=7.9 \times 10^{3} \mathrm{~m} / \mathrm{s}$ $\approx 8 \mathrm{~km} / \mathrm{s}$
Manipal UGET-2019
Gravitation
138699
For a given density of a planet, the orbital speed of satellite near the surface of the planet of radius $R$ is proportional to
1 $R^{1 / 2}$
2 $\mathrm{R}^{3 / 2}$
3 $\mathrm{R}^{-1 / 2}$
4 $\mathrm{R}^{0}$
Explanation:
D From the Kepler's third law - $\frac{\mathrm{T}^{2}}{\mathrm{R}^{3}}=\frac{4 \pi^{2}}{\mathrm{GM}}$ when $\mathrm{r}$ is radius, semi-major axis of the elliptical near the planet surface. $\mathrm{M} =\rho \mathrm{V}$ $=\rho \times \frac{4}{3} \pi \mathrm{R}^{3}$ $\frac{\mathrm{T}^{2}}{\mathrm{R}^{3}} =\frac{4 \pi^{2}}{\mathrm{G} . \rho \times \frac{4}{3} \pi \mathrm{R}^{3}}$ $\mathrm{~T} =\sqrt{\frac{3 \pi}{\mathrm{G} \rho}} \times \mathrm{R}^{0}$ The planet of radius $\mathrm{R}$ is proportional to $\mathrm{R}^{0}$.
138697
The escape velocity from the earth's surface is $11.2 \mathrm{Km} / \mathrm{s}$. If a planet has a radius twice that of the earth and on which the acceleration due to gravity is twice that on the earth, then the escape velocity on this planet will be
138698
A satellite revolves very near to the earth surface. Its speed should be around
1 $5 \mathrm{~km} / \mathrm{s}$
2 $8 \mathrm{~km} / \mathrm{s}$
3 $2 \mathrm{~km} / \mathrm{s}$
4 $11 \mathrm{~km} / \mathrm{s}$
Explanation:
B Since, orbital speed of satellite near to the earth surface is approximately $8 \mathrm{~km} / \mathrm{s}$. We know that, $\text { orbital velocity }(\mathrm{v}) =\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $=\sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6400 \times 1000}}$ $=7.9 \times 10^{3} \mathrm{~m} / \mathrm{s}$ $\approx 8 \mathrm{~km} / \mathrm{s}$
Manipal UGET-2019
Gravitation
138699
For a given density of a planet, the orbital speed of satellite near the surface of the planet of radius $R$ is proportional to
1 $R^{1 / 2}$
2 $\mathrm{R}^{3 / 2}$
3 $\mathrm{R}^{-1 / 2}$
4 $\mathrm{R}^{0}$
Explanation:
D From the Kepler's third law - $\frac{\mathrm{T}^{2}}{\mathrm{R}^{3}}=\frac{4 \pi^{2}}{\mathrm{GM}}$ when $\mathrm{r}$ is radius, semi-major axis of the elliptical near the planet surface. $\mathrm{M} =\rho \mathrm{V}$ $=\rho \times \frac{4}{3} \pi \mathrm{R}^{3}$ $\frac{\mathrm{T}^{2}}{\mathrm{R}^{3}} =\frac{4 \pi^{2}}{\mathrm{G} . \rho \times \frac{4}{3} \pi \mathrm{R}^{3}}$ $\mathrm{~T} =\sqrt{\frac{3 \pi}{\mathrm{G} \rho}} \times \mathrm{R}^{0}$ The planet of radius $\mathrm{R}$ is proportional to $\mathrm{R}^{0}$.
138697
The escape velocity from the earth's surface is $11.2 \mathrm{Km} / \mathrm{s}$. If a planet has a radius twice that of the earth and on which the acceleration due to gravity is twice that on the earth, then the escape velocity on this planet will be
138698
A satellite revolves very near to the earth surface. Its speed should be around
1 $5 \mathrm{~km} / \mathrm{s}$
2 $8 \mathrm{~km} / \mathrm{s}$
3 $2 \mathrm{~km} / \mathrm{s}$
4 $11 \mathrm{~km} / \mathrm{s}$
Explanation:
B Since, orbital speed of satellite near to the earth surface is approximately $8 \mathrm{~km} / \mathrm{s}$. We know that, $\text { orbital velocity }(\mathrm{v}) =\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $=\sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6400 \times 1000}}$ $=7.9 \times 10^{3} \mathrm{~m} / \mathrm{s}$ $\approx 8 \mathrm{~km} / \mathrm{s}$
Manipal UGET-2019
Gravitation
138699
For a given density of a planet, the orbital speed of satellite near the surface of the planet of radius $R$ is proportional to
1 $R^{1 / 2}$
2 $\mathrm{R}^{3 / 2}$
3 $\mathrm{R}^{-1 / 2}$
4 $\mathrm{R}^{0}$
Explanation:
D From the Kepler's third law - $\frac{\mathrm{T}^{2}}{\mathrm{R}^{3}}=\frac{4 \pi^{2}}{\mathrm{GM}}$ when $\mathrm{r}$ is radius, semi-major axis of the elliptical near the planet surface. $\mathrm{M} =\rho \mathrm{V}$ $=\rho \times \frac{4}{3} \pi \mathrm{R}^{3}$ $\frac{\mathrm{T}^{2}}{\mathrm{R}^{3}} =\frac{4 \pi^{2}}{\mathrm{G} . \rho \times \frac{4}{3} \pi \mathrm{R}^{3}}$ $\mathrm{~T} =\sqrt{\frac{3 \pi}{\mathrm{G} \rho}} \times \mathrm{R}^{0}$ The planet of radius $\mathrm{R}$ is proportional to $\mathrm{R}^{0}$.
138697
The escape velocity from the earth's surface is $11.2 \mathrm{Km} / \mathrm{s}$. If a planet has a radius twice that of the earth and on which the acceleration due to gravity is twice that on the earth, then the escape velocity on this planet will be
138698
A satellite revolves very near to the earth surface. Its speed should be around
1 $5 \mathrm{~km} / \mathrm{s}$
2 $8 \mathrm{~km} / \mathrm{s}$
3 $2 \mathrm{~km} / \mathrm{s}$
4 $11 \mathrm{~km} / \mathrm{s}$
Explanation:
B Since, orbital speed of satellite near to the earth surface is approximately $8 \mathrm{~km} / \mathrm{s}$. We know that, $\text { orbital velocity }(\mathrm{v}) =\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $=\sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6400 \times 1000}}$ $=7.9 \times 10^{3} \mathrm{~m} / \mathrm{s}$ $\approx 8 \mathrm{~km} / \mathrm{s}$
Manipal UGET-2019
Gravitation
138699
For a given density of a planet, the orbital speed of satellite near the surface of the planet of radius $R$ is proportional to
1 $R^{1 / 2}$
2 $\mathrm{R}^{3 / 2}$
3 $\mathrm{R}^{-1 / 2}$
4 $\mathrm{R}^{0}$
Explanation:
D From the Kepler's third law - $\frac{\mathrm{T}^{2}}{\mathrm{R}^{3}}=\frac{4 \pi^{2}}{\mathrm{GM}}$ when $\mathrm{r}$ is radius, semi-major axis of the elliptical near the planet surface. $\mathrm{M} =\rho \mathrm{V}$ $=\rho \times \frac{4}{3} \pi \mathrm{R}^{3}$ $\frac{\mathrm{T}^{2}}{\mathrm{R}^{3}} =\frac{4 \pi^{2}}{\mathrm{G} . \rho \times \frac{4}{3} \pi \mathrm{R}^{3}}$ $\mathrm{~T} =\sqrt{\frac{3 \pi}{\mathrm{G} \rho}} \times \mathrm{R}^{0}$ The planet of radius $\mathrm{R}$ is proportional to $\mathrm{R}^{0}$.
