359723
The potential energy of a satellite of mass \(m\) and revolving at a height \(R_{e}\) above the surface of earth where \(R_{e}=\) radius of earth, is
1 \( - mg\,{{\mathop{\rm Re}\nolimits} _e}\)
2 \(\dfrac{-m g R_{e}}{2}\)
3 \(\frac{{ - mg\,{R_e}}}{3}\)
4 \(\frac{{ - mg\,{R_e}}}{4}\)
Explanation:
At a height \(h\) above the surface of earth the gravitational potential energy of the particle of mass \(m\) is \({U_h} = - \frac{{G{M_e}m}}{{{R_e} + h}}\) Where \({M_e}\,\,\& \,\,{R_e}\) are the mass & radius of earth respectively. In this question, since \(h = {R_e}\) So \(U_{h=R_{e}}=-\dfrac{G M_{e} m}{2 R_{e}}=\dfrac{-m g R_{e}}{2}\)
PHXI08:GRAVITATION
359724
Two satellites \(A\) and \(B\) move round the earth in the same orbit. The mass of \(A\) is twice the mass of \(B\). The quantity which is same for the two satellites will be
1 speed
2 potential energy
3 total energy
4 kinetic energy
Explanation:
Orbital speed, \(v_{0}=\sqrt{\dfrac{G M}{R}}\) It is independent of the mass of body. So, the speed will be same for both the satellites \(T E=\dfrac{P E}{2}=-K E=-\dfrac{G M m}{2 r}\) All energies depend on the mass of particle. So, correct option is (1).
JEE - 2023
PHXI08:GRAVITATION
359725
Assertion : If an earth satellite moves to a lower orbit, there is some dissipation of energy but the satellite speed increases. Reason : The speed of satellite is a constant quantity.
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:
Atmospheric resistance causes a satellite to loose kinetic energy, leading it to shift to a lower orbit where increased kinetic energy correspond\(s\) to higher speed. So correct option is (3).
PHXI08:GRAVITATION
359726
A satellite of mass \(m\) is in a circular orbit of radius \(2 R_{E}\) about the earth. The energy required to transfer it to a circular orbit of radius \(4 R_{E}\) is (where \(M_{E}\) and \(R_{E}\) is the mass and radius of the earth respectively):
1 \(\dfrac{G M_{E} m}{2 R_{E}}\)
2 \(\dfrac{G M_{E} m}{4 R_{E}}\)
3 \(\dfrac{G M_{E} m}{8 R_{E}}\)
4 \(\dfrac{G M_{E} m}{16 R_{E}}\)
Explanation:
Initial total energy of the satellite is \(E_{i}=-\dfrac{G M_{E} m}{4 R_{E}}\) Final total energy of the satellite is \(E_{f}=-\dfrac{G M_{E} m}{8 R_{E}}\) The change in the total energy is \(\Delta E=E f-E i\), \(\begin{aligned}& \Delta E=-\dfrac{G M_{E} m}{8 R_{E}}-\left(-\dfrac{G M_{E} m}{4 R_{E}}\right) \\& =-\dfrac{G M_{E} m}{8 R_{E}}+\dfrac{G M_{E} m}{4 R_{E}}=\dfrac{G M_{E} m}{8 R_{E}}\end{aligned}\) Thus, the energy required to transfer the satellite to the desired orbit is \(\dfrac{G M_{E} m}{8 R_{E}}\).
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PHXI08:GRAVITATION
359723
The potential energy of a satellite of mass \(m\) and revolving at a height \(R_{e}\) above the surface of earth where \(R_{e}=\) radius of earth, is
1 \( - mg\,{{\mathop{\rm Re}\nolimits} _e}\)
2 \(\dfrac{-m g R_{e}}{2}\)
3 \(\frac{{ - mg\,{R_e}}}{3}\)
4 \(\frac{{ - mg\,{R_e}}}{4}\)
Explanation:
At a height \(h\) above the surface of earth the gravitational potential energy of the particle of mass \(m\) is \({U_h} = - \frac{{G{M_e}m}}{{{R_e} + h}}\) Where \({M_e}\,\,\& \,\,{R_e}\) are the mass & radius of earth respectively. In this question, since \(h = {R_e}\) So \(U_{h=R_{e}}=-\dfrac{G M_{e} m}{2 R_{e}}=\dfrac{-m g R_{e}}{2}\)
PHXI08:GRAVITATION
359724
Two satellites \(A\) and \(B\) move round the earth in the same orbit. The mass of \(A\) is twice the mass of \(B\). The quantity which is same for the two satellites will be
1 speed
2 potential energy
3 total energy
4 kinetic energy
Explanation:
Orbital speed, \(v_{0}=\sqrt{\dfrac{G M}{R}}\) It is independent of the mass of body. So, the speed will be same for both the satellites \(T E=\dfrac{P E}{2}=-K E=-\dfrac{G M m}{2 r}\) All energies depend on the mass of particle. So, correct option is (1).
JEE - 2023
PHXI08:GRAVITATION
359725
Assertion : If an earth satellite moves to a lower orbit, there is some dissipation of energy but the satellite speed increases. Reason : The speed of satellite is a constant quantity.
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:
Atmospheric resistance causes a satellite to loose kinetic energy, leading it to shift to a lower orbit where increased kinetic energy correspond\(s\) to higher speed. So correct option is (3).
PHXI08:GRAVITATION
359726
A satellite of mass \(m\) is in a circular orbit of radius \(2 R_{E}\) about the earth. The energy required to transfer it to a circular orbit of radius \(4 R_{E}\) is (where \(M_{E}\) and \(R_{E}\) is the mass and radius of the earth respectively):
1 \(\dfrac{G M_{E} m}{2 R_{E}}\)
2 \(\dfrac{G M_{E} m}{4 R_{E}}\)
3 \(\dfrac{G M_{E} m}{8 R_{E}}\)
4 \(\dfrac{G M_{E} m}{16 R_{E}}\)
Explanation:
Initial total energy of the satellite is \(E_{i}=-\dfrac{G M_{E} m}{4 R_{E}}\) Final total energy of the satellite is \(E_{f}=-\dfrac{G M_{E} m}{8 R_{E}}\) The change in the total energy is \(\Delta E=E f-E i\), \(\begin{aligned}& \Delta E=-\dfrac{G M_{E} m}{8 R_{E}}-\left(-\dfrac{G M_{E} m}{4 R_{E}}\right) \\& =-\dfrac{G M_{E} m}{8 R_{E}}+\dfrac{G M_{E} m}{4 R_{E}}=\dfrac{G M_{E} m}{8 R_{E}}\end{aligned}\) Thus, the energy required to transfer the satellite to the desired orbit is \(\dfrac{G M_{E} m}{8 R_{E}}\).
359723
The potential energy of a satellite of mass \(m\) and revolving at a height \(R_{e}\) above the surface of earth where \(R_{e}=\) radius of earth, is
1 \( - mg\,{{\mathop{\rm Re}\nolimits} _e}\)
2 \(\dfrac{-m g R_{e}}{2}\)
3 \(\frac{{ - mg\,{R_e}}}{3}\)
4 \(\frac{{ - mg\,{R_e}}}{4}\)
Explanation:
At a height \(h\) above the surface of earth the gravitational potential energy of the particle of mass \(m\) is \({U_h} = - \frac{{G{M_e}m}}{{{R_e} + h}}\) Where \({M_e}\,\,\& \,\,{R_e}\) are the mass & radius of earth respectively. In this question, since \(h = {R_e}\) So \(U_{h=R_{e}}=-\dfrac{G M_{e} m}{2 R_{e}}=\dfrac{-m g R_{e}}{2}\)
PHXI08:GRAVITATION
359724
Two satellites \(A\) and \(B\) move round the earth in the same orbit. The mass of \(A\) is twice the mass of \(B\). The quantity which is same for the two satellites will be
1 speed
2 potential energy
3 total energy
4 kinetic energy
Explanation:
Orbital speed, \(v_{0}=\sqrt{\dfrac{G M}{R}}\) It is independent of the mass of body. So, the speed will be same for both the satellites \(T E=\dfrac{P E}{2}=-K E=-\dfrac{G M m}{2 r}\) All energies depend on the mass of particle. So, correct option is (1).
JEE - 2023
PHXI08:GRAVITATION
359725
Assertion : If an earth satellite moves to a lower orbit, there is some dissipation of energy but the satellite speed increases. Reason : The speed of satellite is a constant quantity.
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:
Atmospheric resistance causes a satellite to loose kinetic energy, leading it to shift to a lower orbit where increased kinetic energy correspond\(s\) to higher speed. So correct option is (3).
PHXI08:GRAVITATION
359726
A satellite of mass \(m\) is in a circular orbit of radius \(2 R_{E}\) about the earth. The energy required to transfer it to a circular orbit of radius \(4 R_{E}\) is (where \(M_{E}\) and \(R_{E}\) is the mass and radius of the earth respectively):
1 \(\dfrac{G M_{E} m}{2 R_{E}}\)
2 \(\dfrac{G M_{E} m}{4 R_{E}}\)
3 \(\dfrac{G M_{E} m}{8 R_{E}}\)
4 \(\dfrac{G M_{E} m}{16 R_{E}}\)
Explanation:
Initial total energy of the satellite is \(E_{i}=-\dfrac{G M_{E} m}{4 R_{E}}\) Final total energy of the satellite is \(E_{f}=-\dfrac{G M_{E} m}{8 R_{E}}\) The change in the total energy is \(\Delta E=E f-E i\), \(\begin{aligned}& \Delta E=-\dfrac{G M_{E} m}{8 R_{E}}-\left(-\dfrac{G M_{E} m}{4 R_{E}}\right) \\& =-\dfrac{G M_{E} m}{8 R_{E}}+\dfrac{G M_{E} m}{4 R_{E}}=\dfrac{G M_{E} m}{8 R_{E}}\end{aligned}\) Thus, the energy required to transfer the satellite to the desired orbit is \(\dfrac{G M_{E} m}{8 R_{E}}\).
359723
The potential energy of a satellite of mass \(m\) and revolving at a height \(R_{e}\) above the surface of earth where \(R_{e}=\) radius of earth, is
1 \( - mg\,{{\mathop{\rm Re}\nolimits} _e}\)
2 \(\dfrac{-m g R_{e}}{2}\)
3 \(\frac{{ - mg\,{R_e}}}{3}\)
4 \(\frac{{ - mg\,{R_e}}}{4}\)
Explanation:
At a height \(h\) above the surface of earth the gravitational potential energy of the particle of mass \(m\) is \({U_h} = - \frac{{G{M_e}m}}{{{R_e} + h}}\) Where \({M_e}\,\,\& \,\,{R_e}\) are the mass & radius of earth respectively. In this question, since \(h = {R_e}\) So \(U_{h=R_{e}}=-\dfrac{G M_{e} m}{2 R_{e}}=\dfrac{-m g R_{e}}{2}\)
PHXI08:GRAVITATION
359724
Two satellites \(A\) and \(B\) move round the earth in the same orbit. The mass of \(A\) is twice the mass of \(B\). The quantity which is same for the two satellites will be
1 speed
2 potential energy
3 total energy
4 kinetic energy
Explanation:
Orbital speed, \(v_{0}=\sqrt{\dfrac{G M}{R}}\) It is independent of the mass of body. So, the speed will be same for both the satellites \(T E=\dfrac{P E}{2}=-K E=-\dfrac{G M m}{2 r}\) All energies depend on the mass of particle. So, correct option is (1).
JEE - 2023
PHXI08:GRAVITATION
359725
Assertion : If an earth satellite moves to a lower orbit, there is some dissipation of energy but the satellite speed increases. Reason : The speed of satellite is a constant quantity.
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:
Atmospheric resistance causes a satellite to loose kinetic energy, leading it to shift to a lower orbit where increased kinetic energy correspond\(s\) to higher speed. So correct option is (3).
PHXI08:GRAVITATION
359726
A satellite of mass \(m\) is in a circular orbit of radius \(2 R_{E}\) about the earth. The energy required to transfer it to a circular orbit of radius \(4 R_{E}\) is (where \(M_{E}\) and \(R_{E}\) is the mass and radius of the earth respectively):
1 \(\dfrac{G M_{E} m}{2 R_{E}}\)
2 \(\dfrac{G M_{E} m}{4 R_{E}}\)
3 \(\dfrac{G M_{E} m}{8 R_{E}}\)
4 \(\dfrac{G M_{E} m}{16 R_{E}}\)
Explanation:
Initial total energy of the satellite is \(E_{i}=-\dfrac{G M_{E} m}{4 R_{E}}\) Final total energy of the satellite is \(E_{f}=-\dfrac{G M_{E} m}{8 R_{E}}\) The change in the total energy is \(\Delta E=E f-E i\), \(\begin{aligned}& \Delta E=-\dfrac{G M_{E} m}{8 R_{E}}-\left(-\dfrac{G M_{E} m}{4 R_{E}}\right) \\& =-\dfrac{G M_{E} m}{8 R_{E}}+\dfrac{G M_{E} m}{4 R_{E}}=\dfrac{G M_{E} m}{8 R_{E}}\end{aligned}\) Thus, the energy required to transfer the satellite to the desired orbit is \(\dfrac{G M_{E} m}{8 R_{E}}\).
