359706
The additional kinetic energy to be provided to a satellite of mass \(m\) revolving around a planet of mass \(M\), to transfer it from a circular orbit of radius \(R_{1}\) to another of radius \(R_{2}\left(R_{2}>R_{1}\right)\) is
2 \(G m M\left(\dfrac{1}{R_{1}^{2}}-\dfrac{1}{R_{2}^{2}}\right)\)
3 \(\dfrac{1}{2} G m M\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)\)
4 \(2 G m M\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)\)
Explanation:
The total energy of a satellite present in an orbit of radius \(r\) is \(T \cdot E=\dfrac{-G M m}{2 r}\) The difference in the energies of two orbits is equal to the additional K.E to be supplied \(K \cdot E=\dfrac{G M m}{2}\left[\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right]\)
PHXI08:GRAVITATION
359707
Match the Column I and Column II
(Where \(a = \) radius of planet orbit, \(r = \) radius of planet, \(M = \) mass of Sun, \(m = \) mass of planet). Choose the correct answer from the options given below.
1 A - P, B - Q, C - R, D - S
2 A - P, B - S, C - Q, D - R
3 A - R, B - S, C - P, D - Q
4 A - Q, B - P, C - S, D - R
Explanation:
Kinetic energy of a planet in a circular path, \(K=\dfrac{G M m}{2 a}\) \(\therefore(A) \rightarrow(Q)\) Potential energy of sun-planet system, \(U=\dfrac{-G M m}{a}\) ; \(\therefore \quad(B) \rightarrow(P)\) Total energy of planet, \(E=K+U=\dfrac{-G M m}{2 a}\) i.e., \(E=-K=\dfrac{U}{2}\) \(\therefore(C) \rightarrow(S)\) Escape energy at the surface of planet for unit mass object, \(E_{e}=\dfrac{G M}{r}\) \((D) \rightarrow(R)\) So, \(A-Q, B-P, C-S, D-R\), option (4) is correct.
JEE - 2024
PHXI08:GRAVITATION
359708
What is the minimum energy required to launch a satellite of mass \(m\) from the surface of a planet of mass \(M\) and radius \(R\) in a circular orbit at an altitude of \(2 R\) ?
1 \(\dfrac{5 G m M}{6 R}\)
2 \(\dfrac{G m M}{3 R}\)
3 \(\dfrac{2 G m M}{3 R}\)
4 \(\dfrac{G m M}{2 R}\)
Explanation:
As minimum energy is required so the \(K . E_{f}\) is zero. \(\begin{aligned}& T \cdot E_{i}=T \cdot E_{f} \\& U_{i}+K \cdot E_{i}=U_{f}+K \cdot E_{f} \\& \dfrac{-G M m}{r_{1}}+\dfrac{1}{2} m v^{2}=\dfrac{-G M m}{2 r_{2}} \\& r_{1}=R, r_{2}=3 R\end{aligned}\) where \(\dfrac{1}{2} m v^{2}\) - is the energy given to the satellite. \(\Rightarrow \dfrac{1}{2} m v^{2}=\dfrac{G M m}{r_{1}}-\dfrac{G M m}{2 r_{2}}=\dfrac{5 G M m}{6 R}\)
PHXI08:GRAVITATION
359709
The minimum energy required to launch a satellite of mass \({m}\) from the surface of earth of mass \({M}\) and radius \({R}\) in a circular orbit at an altitude of \({2 R}\) from the surface of the earth is:
1 \({\dfrac{5 G m M}{6 R}}\)
2 \({\dfrac{2 G m M}{3 R}}\)
3 \({\dfrac{G m M}{2 R}}\)
4 \({\dfrac{G m M}{3 R}}\)
Explanation:
From conservation of energy \({U_i} + {K_i} = {U_f} + {K_f}\) \( \Rightarrow - \frac{{GMm}}{R} + {K_i} = - \frac{{GMm}}{{3R}} + \frac{1}{2}mv_0^2\) where \({v_{0}}\) is the orbital speed of the satellite present in the orbit of radius \({3 R}\). \({\Rightarrow-\dfrac{G M m}{R}+K_{i}=-\dfrac{G M m}{3 R}+\dfrac{1}{2} \times m \times \dfrac{G M}{3 R}}\) \({\Rightarrow \quad K_{i}=-\dfrac{1}{6} \dfrac{G M m}{R}+\dfrac{G M m}{R}}\) \({K_{i}=\dfrac{5}{6} \dfrac{G M m}{R}}\)
359706
The additional kinetic energy to be provided to a satellite of mass \(m\) revolving around a planet of mass \(M\), to transfer it from a circular orbit of radius \(R_{1}\) to another of radius \(R_{2}\left(R_{2}>R_{1}\right)\) is
2 \(G m M\left(\dfrac{1}{R_{1}^{2}}-\dfrac{1}{R_{2}^{2}}\right)\)
3 \(\dfrac{1}{2} G m M\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)\)
4 \(2 G m M\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)\)
Explanation:
The total energy of a satellite present in an orbit of radius \(r\) is \(T \cdot E=\dfrac{-G M m}{2 r}\) The difference in the energies of two orbits is equal to the additional K.E to be supplied \(K \cdot E=\dfrac{G M m}{2}\left[\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right]\)
PHXI08:GRAVITATION
359707
Match the Column I and Column II
(Where \(a = \) radius of planet orbit, \(r = \) radius of planet, \(M = \) mass of Sun, \(m = \) mass of planet). Choose the correct answer from the options given below.
1 A - P, B - Q, C - R, D - S
2 A - P, B - S, C - Q, D - R
3 A - R, B - S, C - P, D - Q
4 A - Q, B - P, C - S, D - R
Explanation:
Kinetic energy of a planet in a circular path, \(K=\dfrac{G M m}{2 a}\) \(\therefore(A) \rightarrow(Q)\) Potential energy of sun-planet system, \(U=\dfrac{-G M m}{a}\) ; \(\therefore \quad(B) \rightarrow(P)\) Total energy of planet, \(E=K+U=\dfrac{-G M m}{2 a}\) i.e., \(E=-K=\dfrac{U}{2}\) \(\therefore(C) \rightarrow(S)\) Escape energy at the surface of planet for unit mass object, \(E_{e}=\dfrac{G M}{r}\) \((D) \rightarrow(R)\) So, \(A-Q, B-P, C-S, D-R\), option (4) is correct.
JEE - 2024
PHXI08:GRAVITATION
359708
What is the minimum energy required to launch a satellite of mass \(m\) from the surface of a planet of mass \(M\) and radius \(R\) in a circular orbit at an altitude of \(2 R\) ?
1 \(\dfrac{5 G m M}{6 R}\)
2 \(\dfrac{G m M}{3 R}\)
3 \(\dfrac{2 G m M}{3 R}\)
4 \(\dfrac{G m M}{2 R}\)
Explanation:
As minimum energy is required so the \(K . E_{f}\) is zero. \(\begin{aligned}& T \cdot E_{i}=T \cdot E_{f} \\& U_{i}+K \cdot E_{i}=U_{f}+K \cdot E_{f} \\& \dfrac{-G M m}{r_{1}}+\dfrac{1}{2} m v^{2}=\dfrac{-G M m}{2 r_{2}} \\& r_{1}=R, r_{2}=3 R\end{aligned}\) where \(\dfrac{1}{2} m v^{2}\) - is the energy given to the satellite. \(\Rightarrow \dfrac{1}{2} m v^{2}=\dfrac{G M m}{r_{1}}-\dfrac{G M m}{2 r_{2}}=\dfrac{5 G M m}{6 R}\)
PHXI08:GRAVITATION
359709
The minimum energy required to launch a satellite of mass \({m}\) from the surface of earth of mass \({M}\) and radius \({R}\) in a circular orbit at an altitude of \({2 R}\) from the surface of the earth is:
1 \({\dfrac{5 G m M}{6 R}}\)
2 \({\dfrac{2 G m M}{3 R}}\)
3 \({\dfrac{G m M}{2 R}}\)
4 \({\dfrac{G m M}{3 R}}\)
Explanation:
From conservation of energy \({U_i} + {K_i} = {U_f} + {K_f}\) \( \Rightarrow - \frac{{GMm}}{R} + {K_i} = - \frac{{GMm}}{{3R}} + \frac{1}{2}mv_0^2\) where \({v_{0}}\) is the orbital speed of the satellite present in the orbit of radius \({3 R}\). \({\Rightarrow-\dfrac{G M m}{R}+K_{i}=-\dfrac{G M m}{3 R}+\dfrac{1}{2} \times m \times \dfrac{G M}{3 R}}\) \({\Rightarrow \quad K_{i}=-\dfrac{1}{6} \dfrac{G M m}{R}+\dfrac{G M m}{R}}\) \({K_{i}=\dfrac{5}{6} \dfrac{G M m}{R}}\)
359706
The additional kinetic energy to be provided to a satellite of mass \(m\) revolving around a planet of mass \(M\), to transfer it from a circular orbit of radius \(R_{1}\) to another of radius \(R_{2}\left(R_{2}>R_{1}\right)\) is
2 \(G m M\left(\dfrac{1}{R_{1}^{2}}-\dfrac{1}{R_{2}^{2}}\right)\)
3 \(\dfrac{1}{2} G m M\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)\)
4 \(2 G m M\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)\)
Explanation:
The total energy of a satellite present in an orbit of radius \(r\) is \(T \cdot E=\dfrac{-G M m}{2 r}\) The difference in the energies of two orbits is equal to the additional K.E to be supplied \(K \cdot E=\dfrac{G M m}{2}\left[\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right]\)
PHXI08:GRAVITATION
359707
Match the Column I and Column II
(Where \(a = \) radius of planet orbit, \(r = \) radius of planet, \(M = \) mass of Sun, \(m = \) mass of planet). Choose the correct answer from the options given below.
1 A - P, B - Q, C - R, D - S
2 A - P, B - S, C - Q, D - R
3 A - R, B - S, C - P, D - Q
4 A - Q, B - P, C - S, D - R
Explanation:
Kinetic energy of a planet in a circular path, \(K=\dfrac{G M m}{2 a}\) \(\therefore(A) \rightarrow(Q)\) Potential energy of sun-planet system, \(U=\dfrac{-G M m}{a}\) ; \(\therefore \quad(B) \rightarrow(P)\) Total energy of planet, \(E=K+U=\dfrac{-G M m}{2 a}\) i.e., \(E=-K=\dfrac{U}{2}\) \(\therefore(C) \rightarrow(S)\) Escape energy at the surface of planet for unit mass object, \(E_{e}=\dfrac{G M}{r}\) \((D) \rightarrow(R)\) So, \(A-Q, B-P, C-S, D-R\), option (4) is correct.
JEE - 2024
PHXI08:GRAVITATION
359708
What is the minimum energy required to launch a satellite of mass \(m\) from the surface of a planet of mass \(M\) and radius \(R\) in a circular orbit at an altitude of \(2 R\) ?
1 \(\dfrac{5 G m M}{6 R}\)
2 \(\dfrac{G m M}{3 R}\)
3 \(\dfrac{2 G m M}{3 R}\)
4 \(\dfrac{G m M}{2 R}\)
Explanation:
As minimum energy is required so the \(K . E_{f}\) is zero. \(\begin{aligned}& T \cdot E_{i}=T \cdot E_{f} \\& U_{i}+K \cdot E_{i}=U_{f}+K \cdot E_{f} \\& \dfrac{-G M m}{r_{1}}+\dfrac{1}{2} m v^{2}=\dfrac{-G M m}{2 r_{2}} \\& r_{1}=R, r_{2}=3 R\end{aligned}\) where \(\dfrac{1}{2} m v^{2}\) - is the energy given to the satellite. \(\Rightarrow \dfrac{1}{2} m v^{2}=\dfrac{G M m}{r_{1}}-\dfrac{G M m}{2 r_{2}}=\dfrac{5 G M m}{6 R}\)
PHXI08:GRAVITATION
359709
The minimum energy required to launch a satellite of mass \({m}\) from the surface of earth of mass \({M}\) and radius \({R}\) in a circular orbit at an altitude of \({2 R}\) from the surface of the earth is:
1 \({\dfrac{5 G m M}{6 R}}\)
2 \({\dfrac{2 G m M}{3 R}}\)
3 \({\dfrac{G m M}{2 R}}\)
4 \({\dfrac{G m M}{3 R}}\)
Explanation:
From conservation of energy \({U_i} + {K_i} = {U_f} + {K_f}\) \( \Rightarrow - \frac{{GMm}}{R} + {K_i} = - \frac{{GMm}}{{3R}} + \frac{1}{2}mv_0^2\) where \({v_{0}}\) is the orbital speed of the satellite present in the orbit of radius \({3 R}\). \({\Rightarrow-\dfrac{G M m}{R}+K_{i}=-\dfrac{G M m}{3 R}+\dfrac{1}{2} \times m \times \dfrac{G M}{3 R}}\) \({\Rightarrow \quad K_{i}=-\dfrac{1}{6} \dfrac{G M m}{R}+\dfrac{G M m}{R}}\) \({K_{i}=\dfrac{5}{6} \dfrac{G M m}{R}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI08:GRAVITATION
359706
The additional kinetic energy to be provided to a satellite of mass \(m\) revolving around a planet of mass \(M\), to transfer it from a circular orbit of radius \(R_{1}\) to another of radius \(R_{2}\left(R_{2}>R_{1}\right)\) is
2 \(G m M\left(\dfrac{1}{R_{1}^{2}}-\dfrac{1}{R_{2}^{2}}\right)\)
3 \(\dfrac{1}{2} G m M\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)\)
4 \(2 G m M\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)\)
Explanation:
The total energy of a satellite present in an orbit of radius \(r\) is \(T \cdot E=\dfrac{-G M m}{2 r}\) The difference in the energies of two orbits is equal to the additional K.E to be supplied \(K \cdot E=\dfrac{G M m}{2}\left[\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right]\)
PHXI08:GRAVITATION
359707
Match the Column I and Column II
(Where \(a = \) radius of planet orbit, \(r = \) radius of planet, \(M = \) mass of Sun, \(m = \) mass of planet). Choose the correct answer from the options given below.
1 A - P, B - Q, C - R, D - S
2 A - P, B - S, C - Q, D - R
3 A - R, B - S, C - P, D - Q
4 A - Q, B - P, C - S, D - R
Explanation:
Kinetic energy of a planet in a circular path, \(K=\dfrac{G M m}{2 a}\) \(\therefore(A) \rightarrow(Q)\) Potential energy of sun-planet system, \(U=\dfrac{-G M m}{a}\) ; \(\therefore \quad(B) \rightarrow(P)\) Total energy of planet, \(E=K+U=\dfrac{-G M m}{2 a}\) i.e., \(E=-K=\dfrac{U}{2}\) \(\therefore(C) \rightarrow(S)\) Escape energy at the surface of planet for unit mass object, \(E_{e}=\dfrac{G M}{r}\) \((D) \rightarrow(R)\) So, \(A-Q, B-P, C-S, D-R\), option (4) is correct.
JEE - 2024
PHXI08:GRAVITATION
359708
What is the minimum energy required to launch a satellite of mass \(m\) from the surface of a planet of mass \(M\) and radius \(R\) in a circular orbit at an altitude of \(2 R\) ?
1 \(\dfrac{5 G m M}{6 R}\)
2 \(\dfrac{G m M}{3 R}\)
3 \(\dfrac{2 G m M}{3 R}\)
4 \(\dfrac{G m M}{2 R}\)
Explanation:
As minimum energy is required so the \(K . E_{f}\) is zero. \(\begin{aligned}& T \cdot E_{i}=T \cdot E_{f} \\& U_{i}+K \cdot E_{i}=U_{f}+K \cdot E_{f} \\& \dfrac{-G M m}{r_{1}}+\dfrac{1}{2} m v^{2}=\dfrac{-G M m}{2 r_{2}} \\& r_{1}=R, r_{2}=3 R\end{aligned}\) where \(\dfrac{1}{2} m v^{2}\) - is the energy given to the satellite. \(\Rightarrow \dfrac{1}{2} m v^{2}=\dfrac{G M m}{r_{1}}-\dfrac{G M m}{2 r_{2}}=\dfrac{5 G M m}{6 R}\)
PHXI08:GRAVITATION
359709
The minimum energy required to launch a satellite of mass \({m}\) from the surface of earth of mass \({M}\) and radius \({R}\) in a circular orbit at an altitude of \({2 R}\) from the surface of the earth is:
1 \({\dfrac{5 G m M}{6 R}}\)
2 \({\dfrac{2 G m M}{3 R}}\)
3 \({\dfrac{G m M}{2 R}}\)
4 \({\dfrac{G m M}{3 R}}\)
Explanation:
From conservation of energy \({U_i} + {K_i} = {U_f} + {K_f}\) \( \Rightarrow - \frac{{GMm}}{R} + {K_i} = - \frac{{GMm}}{{3R}} + \frac{1}{2}mv_0^2\) where \({v_{0}}\) is the orbital speed of the satellite present in the orbit of radius \({3 R}\). \({\Rightarrow-\dfrac{G M m}{R}+K_{i}=-\dfrac{G M m}{3 R}+\dfrac{1}{2} \times m \times \dfrac{G M}{3 R}}\) \({\Rightarrow \quad K_{i}=-\dfrac{1}{6} \dfrac{G M m}{R}+\dfrac{G M m}{R}}\) \({K_{i}=\dfrac{5}{6} \dfrac{G M m}{R}}\)
