359753
Two satellites of masses \(m\) and \(3 m\) revolve around the earth in circular orbits of radii \(r\) and \(3 r\) respectively. The ratio of orbital speeds of the satellites respectively is
1 \(9: 1\)
2 \(\sqrt{3}: 1\)
3 \(3: 1\)
4 \(1: 1\)
Explanation:
Orbital speed of satellite is, \(v=\sqrt{\dfrac{G M}{r}}\) \(\therefore v \propto \dfrac{1}{\sqrt{r}}\) So, as \(r\) increases, \(v\) decreases Given : \(r_{1}=r, r_{2}=3 r\) \(\Rightarrow \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}=\sqrt{\dfrac{3 r}{r}}=\sqrt{3}: 1\) So, correct option is (2).
JEE - 2023
PHXI08:GRAVITATION
359754
A satellite is moving on a circular path of radius \(r\) around the earth has a time period \(T\). If its radius slightly increased by \(\Delta r\), the change in its time period is:
359755
The time period ' \(T\) ' of a satellite is related to the density \((\rho)\) of the planet which is orbiting close around the planet as
1 \(T \propto \rho^{\frac{-1}{2}}\)
2 \(T \propto \rho\)
3 \(T \propto \rho^{\frac{1}{2}}\)
4 \(T \propto \rho^{\frac{-3}{2}}\)
Explanation:
Time period of satellite is given by, \(T=2 \pi \sqrt{\dfrac{r^{3}}{G M}}\) Where ' \(r\) ' is the radius of the orbit and ' \(M\) ' is the mass of the planet, As the satellite is revolving very close to the planet, so \(r=R\). Also, \(M=\rho \times V=\rho \times \dfrac{4}{3} \pi R^{3}\) \(T=2 \pi \sqrt{\dfrac{R^{3}}{G \times \dfrac{4 \pi}{3} \rho R^{3}}}=\sqrt{\dfrac{3 \pi}{\rho G}} \text { i.e. } T \propto \rho^{-\dfrac{1}{2}}\)
MHTCET - 2022
PHXI08:GRAVITATION
359756
Assertion : The speed of revolution of an artificial satellite revolving very near the earth is \(8 \mathrm{kms}^{-1}\) Reason : Orbital velocity of a satellite, become independent of height of satellite.
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:
Orbital velocity \(v_{0}=\sqrt{R_{E} g}=\sqrt{6400 \times 10^{3} \times 10}\) \( \Rightarrow {v_0} = 8 \times {10^3}\;m{\rm{/}}s = 8\;km{\rm{/}}s\) Here \(\left(R_{E}+h\right) \approx R_{E} \quad\left(\right.\) since \(\left.h < < R_{E}\right)\). Assertion says "very near to surface of earth". The orbital velocity of a satellite depends on the mass of the Earth and the radius of the satellite's orbit, but it is largely independent of the height \(h\) of the satellite above the Earth's surface. (Since \(h\) is small compared to \(R_{E}\) ). So correct option is (1).
359753
Two satellites of masses \(m\) and \(3 m\) revolve around the earth in circular orbits of radii \(r\) and \(3 r\) respectively. The ratio of orbital speeds of the satellites respectively is
1 \(9: 1\)
2 \(\sqrt{3}: 1\)
3 \(3: 1\)
4 \(1: 1\)
Explanation:
Orbital speed of satellite is, \(v=\sqrt{\dfrac{G M}{r}}\) \(\therefore v \propto \dfrac{1}{\sqrt{r}}\) So, as \(r\) increases, \(v\) decreases Given : \(r_{1}=r, r_{2}=3 r\) \(\Rightarrow \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}=\sqrt{\dfrac{3 r}{r}}=\sqrt{3}: 1\) So, correct option is (2).
JEE - 2023
PHXI08:GRAVITATION
359754
A satellite is moving on a circular path of radius \(r\) around the earth has a time period \(T\). If its radius slightly increased by \(\Delta r\), the change in its time period is:
359755
The time period ' \(T\) ' of a satellite is related to the density \((\rho)\) of the planet which is orbiting close around the planet as
1 \(T \propto \rho^{\frac{-1}{2}}\)
2 \(T \propto \rho\)
3 \(T \propto \rho^{\frac{1}{2}}\)
4 \(T \propto \rho^{\frac{-3}{2}}\)
Explanation:
Time period of satellite is given by, \(T=2 \pi \sqrt{\dfrac{r^{3}}{G M}}\) Where ' \(r\) ' is the radius of the orbit and ' \(M\) ' is the mass of the planet, As the satellite is revolving very close to the planet, so \(r=R\). Also, \(M=\rho \times V=\rho \times \dfrac{4}{3} \pi R^{3}\) \(T=2 \pi \sqrt{\dfrac{R^{3}}{G \times \dfrac{4 \pi}{3} \rho R^{3}}}=\sqrt{\dfrac{3 \pi}{\rho G}} \text { i.e. } T \propto \rho^{-\dfrac{1}{2}}\)
MHTCET - 2022
PHXI08:GRAVITATION
359756
Assertion : The speed of revolution of an artificial satellite revolving very near the earth is \(8 \mathrm{kms}^{-1}\) Reason : Orbital velocity of a satellite, become independent of height of satellite.
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:
Orbital velocity \(v_{0}=\sqrt{R_{E} g}=\sqrt{6400 \times 10^{3} \times 10}\) \( \Rightarrow {v_0} = 8 \times {10^3}\;m{\rm{/}}s = 8\;km{\rm{/}}s\) Here \(\left(R_{E}+h\right) \approx R_{E} \quad\left(\right.\) since \(\left.h < < R_{E}\right)\). Assertion says "very near to surface of earth". The orbital velocity of a satellite depends on the mass of the Earth and the radius of the satellite's orbit, but it is largely independent of the height \(h\) of the satellite above the Earth's surface. (Since \(h\) is small compared to \(R_{E}\) ). So correct option is (1).
359753
Two satellites of masses \(m\) and \(3 m\) revolve around the earth in circular orbits of radii \(r\) and \(3 r\) respectively. The ratio of orbital speeds of the satellites respectively is
1 \(9: 1\)
2 \(\sqrt{3}: 1\)
3 \(3: 1\)
4 \(1: 1\)
Explanation:
Orbital speed of satellite is, \(v=\sqrt{\dfrac{G M}{r}}\) \(\therefore v \propto \dfrac{1}{\sqrt{r}}\) So, as \(r\) increases, \(v\) decreases Given : \(r_{1}=r, r_{2}=3 r\) \(\Rightarrow \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}=\sqrt{\dfrac{3 r}{r}}=\sqrt{3}: 1\) So, correct option is (2).
JEE - 2023
PHXI08:GRAVITATION
359754
A satellite is moving on a circular path of radius \(r\) around the earth has a time period \(T\). If its radius slightly increased by \(\Delta r\), the change in its time period is:
359755
The time period ' \(T\) ' of a satellite is related to the density \((\rho)\) of the planet which is orbiting close around the planet as
1 \(T \propto \rho^{\frac{-1}{2}}\)
2 \(T \propto \rho\)
3 \(T \propto \rho^{\frac{1}{2}}\)
4 \(T \propto \rho^{\frac{-3}{2}}\)
Explanation:
Time period of satellite is given by, \(T=2 \pi \sqrt{\dfrac{r^{3}}{G M}}\) Where ' \(r\) ' is the radius of the orbit and ' \(M\) ' is the mass of the planet, As the satellite is revolving very close to the planet, so \(r=R\). Also, \(M=\rho \times V=\rho \times \dfrac{4}{3} \pi R^{3}\) \(T=2 \pi \sqrt{\dfrac{R^{3}}{G \times \dfrac{4 \pi}{3} \rho R^{3}}}=\sqrt{\dfrac{3 \pi}{\rho G}} \text { i.e. } T \propto \rho^{-\dfrac{1}{2}}\)
MHTCET - 2022
PHXI08:GRAVITATION
359756
Assertion : The speed of revolution of an artificial satellite revolving very near the earth is \(8 \mathrm{kms}^{-1}\) Reason : Orbital velocity of a satellite, become independent of height of satellite.
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:
Orbital velocity \(v_{0}=\sqrt{R_{E} g}=\sqrt{6400 \times 10^{3} \times 10}\) \( \Rightarrow {v_0} = 8 \times {10^3}\;m{\rm{/}}s = 8\;km{\rm{/}}s\) Here \(\left(R_{E}+h\right) \approx R_{E} \quad\left(\right.\) since \(\left.h < < R_{E}\right)\). Assertion says "very near to surface of earth". The orbital velocity of a satellite depends on the mass of the Earth and the radius of the satellite's orbit, but it is largely independent of the height \(h\) of the satellite above the Earth's surface. (Since \(h\) is small compared to \(R_{E}\) ). So correct option is (1).
359753
Two satellites of masses \(m\) and \(3 m\) revolve around the earth in circular orbits of radii \(r\) and \(3 r\) respectively. The ratio of orbital speeds of the satellites respectively is
1 \(9: 1\)
2 \(\sqrt{3}: 1\)
3 \(3: 1\)
4 \(1: 1\)
Explanation:
Orbital speed of satellite is, \(v=\sqrt{\dfrac{G M}{r}}\) \(\therefore v \propto \dfrac{1}{\sqrt{r}}\) So, as \(r\) increases, \(v\) decreases Given : \(r_{1}=r, r_{2}=3 r\) \(\Rightarrow \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}=\sqrt{\dfrac{3 r}{r}}=\sqrt{3}: 1\) So, correct option is (2).
JEE - 2023
PHXI08:GRAVITATION
359754
A satellite is moving on a circular path of radius \(r\) around the earth has a time period \(T\). If its radius slightly increased by \(\Delta r\), the change in its time period is:
359755
The time period ' \(T\) ' of a satellite is related to the density \((\rho)\) of the planet which is orbiting close around the planet as
1 \(T \propto \rho^{\frac{-1}{2}}\)
2 \(T \propto \rho\)
3 \(T \propto \rho^{\frac{1}{2}}\)
4 \(T \propto \rho^{\frac{-3}{2}}\)
Explanation:
Time period of satellite is given by, \(T=2 \pi \sqrt{\dfrac{r^{3}}{G M}}\) Where ' \(r\) ' is the radius of the orbit and ' \(M\) ' is the mass of the planet, As the satellite is revolving very close to the planet, so \(r=R\). Also, \(M=\rho \times V=\rho \times \dfrac{4}{3} \pi R^{3}\) \(T=2 \pi \sqrt{\dfrac{R^{3}}{G \times \dfrac{4 \pi}{3} \rho R^{3}}}=\sqrt{\dfrac{3 \pi}{\rho G}} \text { i.e. } T \propto \rho^{-\dfrac{1}{2}}\)
MHTCET - 2022
PHXI08:GRAVITATION
359756
Assertion : The speed of revolution of an artificial satellite revolving very near the earth is \(8 \mathrm{kms}^{-1}\) Reason : Orbital velocity of a satellite, become independent of height of satellite.
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:
Orbital velocity \(v_{0}=\sqrt{R_{E} g}=\sqrt{6400 \times 10^{3} \times 10}\) \( \Rightarrow {v_0} = 8 \times {10^3}\;m{\rm{/}}s = 8\;km{\rm{/}}s\) Here \(\left(R_{E}+h\right) \approx R_{E} \quad\left(\right.\) since \(\left.h < < R_{E}\right)\). Assertion says "very near to surface of earth". The orbital velocity of a satellite depends on the mass of the Earth and the radius of the satellite's orbit, but it is largely independent of the height \(h\) of the satellite above the Earth's surface. (Since \(h\) is small compared to \(R_{E}\) ). So correct option is (1).
