359775
What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth. Given : (i) The value of gravitational acceleration \(g = 10\;m\;{s^{ - 2}}\) (ii) Radius of earth \({R_E} = 6400\;km.\) Take \(\pi=3.14\)
1 83.73 minutes
2 85 minutes
3 90 minutes
4 156 minutes
Explanation:
Period of revolution of earth satellite is \(T = 2\pi \sqrt {\frac{{{{\left( {{R_E} + h} \right)}^3}}}{{gR_E^2}}} \) As \(h < < R_{E}\) \(T=2 \pi \sqrt{\dfrac{R_{E}}{g}}\) Substituting the given values, we get \(T = 2 \times 3.14\sqrt {\frac{{6400 \times {{10}^3}}}{{10}}} \;s\) \( = 5024\;s = 83.73\) minutes
KCET - 2011
PHXI08:GRAVITATION
359776
If \(g \propto \dfrac{1}{R^{3}}\left(\right.\) instead of \(\frac{1}{{{R^2}}}\)), then the relation between time period of a satellite near earth's surface and radius \(R\) will be
1 \(T \propto R^{2}\)
2 \(T \propto T\)
3 \(T^{2} \propto R\)
4 \(T^{2} \propto R^{2}\)
Explanation:
Gravitational force provides the required centripetal force \(\begin{aligned}& m \omega^{2} R=m g \\& m \omega^{2} R=\dfrac{G M m}{R^{3}} \Rightarrow \dfrac{4 \pi^{2}}{T^{2}}=\dfrac{G M}{R^{4}} \Rightarrow T \propto R^{2} .\end{aligned}\)
PHXI08:GRAVITATION
359777
The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is \(v\). For a satellite orbiting at an altitude of half the earth's radius the orbital velocity is
359778
Assertion : Escape velocity is greater than its orbital velocity of a satellite. Reason : Orbit of a satellite is within the gravitational field of earth whereas escaping is beyond the gravitational field of earth.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(v_{e}=\sqrt{2} v_{0} \Rightarrow v_{e}>v_{0}\) \({v_e} = 11.2\;km{\rm{/}}s\) So correct option is (1).
359775
What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth. Given : (i) The value of gravitational acceleration \(g = 10\;m\;{s^{ - 2}}\) (ii) Radius of earth \({R_E} = 6400\;km.\) Take \(\pi=3.14\)
1 83.73 minutes
2 85 minutes
3 90 minutes
4 156 minutes
Explanation:
Period of revolution of earth satellite is \(T = 2\pi \sqrt {\frac{{{{\left( {{R_E} + h} \right)}^3}}}{{gR_E^2}}} \) As \(h < < R_{E}\) \(T=2 \pi \sqrt{\dfrac{R_{E}}{g}}\) Substituting the given values, we get \(T = 2 \times 3.14\sqrt {\frac{{6400 \times {{10}^3}}}{{10}}} \;s\) \( = 5024\;s = 83.73\) minutes
KCET - 2011
PHXI08:GRAVITATION
359776
If \(g \propto \dfrac{1}{R^{3}}\left(\right.\) instead of \(\frac{1}{{{R^2}}}\)), then the relation between time period of a satellite near earth's surface and radius \(R\) will be
1 \(T \propto R^{2}\)
2 \(T \propto T\)
3 \(T^{2} \propto R\)
4 \(T^{2} \propto R^{2}\)
Explanation:
Gravitational force provides the required centripetal force \(\begin{aligned}& m \omega^{2} R=m g \\& m \omega^{2} R=\dfrac{G M m}{R^{3}} \Rightarrow \dfrac{4 \pi^{2}}{T^{2}}=\dfrac{G M}{R^{4}} \Rightarrow T \propto R^{2} .\end{aligned}\)
PHXI08:GRAVITATION
359777
The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is \(v\). For a satellite orbiting at an altitude of half the earth's radius the orbital velocity is
359778
Assertion : Escape velocity is greater than its orbital velocity of a satellite. Reason : Orbit of a satellite is within the gravitational field of earth whereas escaping is beyond the gravitational field of earth.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(v_{e}=\sqrt{2} v_{0} \Rightarrow v_{e}>v_{0}\) \({v_e} = 11.2\;km{\rm{/}}s\) So correct option is (1).
359775
What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth. Given : (i) The value of gravitational acceleration \(g = 10\;m\;{s^{ - 2}}\) (ii) Radius of earth \({R_E} = 6400\;km.\) Take \(\pi=3.14\)
1 83.73 minutes
2 85 minutes
3 90 minutes
4 156 minutes
Explanation:
Period of revolution of earth satellite is \(T = 2\pi \sqrt {\frac{{{{\left( {{R_E} + h} \right)}^3}}}{{gR_E^2}}} \) As \(h < < R_{E}\) \(T=2 \pi \sqrt{\dfrac{R_{E}}{g}}\) Substituting the given values, we get \(T = 2 \times 3.14\sqrt {\frac{{6400 \times {{10}^3}}}{{10}}} \;s\) \( = 5024\;s = 83.73\) minutes
KCET - 2011
PHXI08:GRAVITATION
359776
If \(g \propto \dfrac{1}{R^{3}}\left(\right.\) instead of \(\frac{1}{{{R^2}}}\)), then the relation between time period of a satellite near earth's surface and radius \(R\) will be
1 \(T \propto R^{2}\)
2 \(T \propto T\)
3 \(T^{2} \propto R\)
4 \(T^{2} \propto R^{2}\)
Explanation:
Gravitational force provides the required centripetal force \(\begin{aligned}& m \omega^{2} R=m g \\& m \omega^{2} R=\dfrac{G M m}{R^{3}} \Rightarrow \dfrac{4 \pi^{2}}{T^{2}}=\dfrac{G M}{R^{4}} \Rightarrow T \propto R^{2} .\end{aligned}\)
PHXI08:GRAVITATION
359777
The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is \(v\). For a satellite orbiting at an altitude of half the earth's radius the orbital velocity is
359778
Assertion : Escape velocity is greater than its orbital velocity of a satellite. Reason : Orbit of a satellite is within the gravitational field of earth whereas escaping is beyond the gravitational field of earth.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(v_{e}=\sqrt{2} v_{0} \Rightarrow v_{e}>v_{0}\) \({v_e} = 11.2\;km{\rm{/}}s\) So correct option is (1).
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI08:GRAVITATION
359775
What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth. Given : (i) The value of gravitational acceleration \(g = 10\;m\;{s^{ - 2}}\) (ii) Radius of earth \({R_E} = 6400\;km.\) Take \(\pi=3.14\)
1 83.73 minutes
2 85 minutes
3 90 minutes
4 156 minutes
Explanation:
Period of revolution of earth satellite is \(T = 2\pi \sqrt {\frac{{{{\left( {{R_E} + h} \right)}^3}}}{{gR_E^2}}} \) As \(h < < R_{E}\) \(T=2 \pi \sqrt{\dfrac{R_{E}}{g}}\) Substituting the given values, we get \(T = 2 \times 3.14\sqrt {\frac{{6400 \times {{10}^3}}}{{10}}} \;s\) \( = 5024\;s = 83.73\) minutes
KCET - 2011
PHXI08:GRAVITATION
359776
If \(g \propto \dfrac{1}{R^{3}}\left(\right.\) instead of \(\frac{1}{{{R^2}}}\)), then the relation between time period of a satellite near earth's surface and radius \(R\) will be
1 \(T \propto R^{2}\)
2 \(T \propto T\)
3 \(T^{2} \propto R\)
4 \(T^{2} \propto R^{2}\)
Explanation:
Gravitational force provides the required centripetal force \(\begin{aligned}& m \omega^{2} R=m g \\& m \omega^{2} R=\dfrac{G M m}{R^{3}} \Rightarrow \dfrac{4 \pi^{2}}{T^{2}}=\dfrac{G M}{R^{4}} \Rightarrow T \propto R^{2} .\end{aligned}\)
PHXI08:GRAVITATION
359777
The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is \(v\). For a satellite orbiting at an altitude of half the earth's radius the orbital velocity is
359778
Assertion : Escape velocity is greater than its orbital velocity of a satellite. Reason : Orbit of a satellite is within the gravitational field of earth whereas escaping is beyond the gravitational field of earth.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(v_{e}=\sqrt{2} v_{0} \Rightarrow v_{e}>v_{0}\) \({v_e} = 11.2\;km{\rm{/}}s\) So correct option is (1).
