Earth Satellites
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PHXI08:GRAVITATION

359697 A satellite is orbiting close to the earth and has a kinetic energy \(K\). The minimum extra kinetic energy required by it to just overcome the gravitation pull of the earth is

1 \(\sqrt 3 \;K\)
2 \(K\)
3 \(2\sqrt 2 \;K\)
4 \(2K\)
KCET - 2019
PHXI08:GRAVITATION

359698 A satellite orbits the earth at a height of 400 \(km\) above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite \(=200 kg\), mass of the earth \( = 6.0 \times {10^{24}}\;kg\), radius of the earth \( = 6.4 \times {10^6}\;m\),
\(G = 6.67 \times {10^{ - 11}}\;N - {m^2}/k{g^2}\)

1 \(5.2 \times {10^{10}}\;J\)
2 \(3 \times {10^6}\;J\)
3 \(4 \times {10^6}\;J\)
4 \(6 \times {10^9}\;J\)
NCERT Exemplar
PHXI08:GRAVITATION

359699 Assertion :
The binding energy of a satellite does not depend upon the mass of the satellite.
Reason :
Binding energy is the negative value of total energy 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.
PHXI08:GRAVITATION

359700 The binding energy of a satellite of mass \(m\) in an orbit of radius \(r\) is (\(R=\) radius of earth, \(g=\) acceleration due to gravity):

1 \(\dfrac{m g R^{2}}{2 r}\)
2 \(\dfrac{m g R^{2}}{r}\)
3 \(-\dfrac{m g R^{2}}{2 r}\)
4 \(-\dfrac{m g R^{2}}{r}\)
PHXI08:GRAVITATION

359701 The ratio of binding energy of a statellite at rest on earth's surface to the binding energy of a statellte of same mass revolving around the earth at a height \(h\) above the earth's surface is (\(R=\) radius of earth)

1 \(\dfrac{2(R+h)}{R}\)
2 \(\dfrac{R+h}{2}\)
3 \(\dfrac{R+h}{R}\)
4 \(\dfrac{R}{R+h}\)
MHTCET - 2020
NEET DIGITAL LIBRARY
PHXI08:GRAVITATION

359697 A satellite is orbiting close to the earth and has a kinetic energy \(K\). The minimum extra kinetic energy required by it to just overcome the gravitation pull of the earth is

1 \(\sqrt 3 \;K\)
2 \(K\)
3 \(2\sqrt 2 \;K\)
4 \(2K\)
KCET - 2019
PHXI08:GRAVITATION

359698 A satellite orbits the earth at a height of 400 \(km\) above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite \(=200 kg\), mass of the earth \( = 6.0 \times {10^{24}}\;kg\), radius of the earth \( = 6.4 \times {10^6}\;m\),
\(G = 6.67 \times {10^{ - 11}}\;N - {m^2}/k{g^2}\)

1 \(5.2 \times {10^{10}}\;J\)
2 \(3 \times {10^6}\;J\)
3 \(4 \times {10^6}\;J\)
4 \(6 \times {10^9}\;J\)
NCERT Exemplar
PHXI08:GRAVITATION

359699 Assertion :
The binding energy of a satellite does not depend upon the mass of the satellite.
Reason :
Binding energy is the negative value of total energy 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.
PHXI08:GRAVITATION

359700 The binding energy of a satellite of mass \(m\) in an orbit of radius \(r\) is (\(R=\) radius of earth, \(g=\) acceleration due to gravity):

1 \(\dfrac{m g R^{2}}{2 r}\)
2 \(\dfrac{m g R^{2}}{r}\)
3 \(-\dfrac{m g R^{2}}{2 r}\)
4 \(-\dfrac{m g R^{2}}{r}\)
PHXI08:GRAVITATION

359701 The ratio of binding energy of a statellite at rest on earth's surface to the binding energy of a statellte of same mass revolving around the earth at a height \(h\) above the earth's surface is (\(R=\) radius of earth)

1 \(\dfrac{2(R+h)}{R}\)
2 \(\dfrac{R+h}{2}\)
3 \(\dfrac{R+h}{R}\)
4 \(\dfrac{R}{R+h}\)
MHTCET - 2020
Join With Us for More Updates
PHXI08:GRAVITATION

359697 A satellite is orbiting close to the earth and has a kinetic energy \(K\). The minimum extra kinetic energy required by it to just overcome the gravitation pull of the earth is

1 \(\sqrt 3 \;K\)
2 \(K\)
3 \(2\sqrt 2 \;K\)
4 \(2K\)
KCET - 2019
PHXI08:GRAVITATION

359698 A satellite orbits the earth at a height of 400 \(km\) above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite \(=200 kg\), mass of the earth \( = 6.0 \times {10^{24}}\;kg\), radius of the earth \( = 6.4 \times {10^6}\;m\),
\(G = 6.67 \times {10^{ - 11}}\;N - {m^2}/k{g^2}\)

1 \(5.2 \times {10^{10}}\;J\)
2 \(3 \times {10^6}\;J\)
3 \(4 \times {10^6}\;J\)
4 \(6 \times {10^9}\;J\)
NCERT Exemplar
PHXI08:GRAVITATION

359699 Assertion :
The binding energy of a satellite does not depend upon the mass of the satellite.
Reason :
Binding energy is the negative value of total energy 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.
PHXI08:GRAVITATION

359700 The binding energy of a satellite of mass \(m\) in an orbit of radius \(r\) is (\(R=\) radius of earth, \(g=\) acceleration due to gravity):

1 \(\dfrac{m g R^{2}}{2 r}\)
2 \(\dfrac{m g R^{2}}{r}\)
3 \(-\dfrac{m g R^{2}}{2 r}\)
4 \(-\dfrac{m g R^{2}}{r}\)
PHXI08:GRAVITATION

359701 The ratio of binding energy of a statellite at rest on earth's surface to the binding energy of a statellte of same mass revolving around the earth at a height \(h\) above the earth's surface is (\(R=\) radius of earth)

1 \(\dfrac{2(R+h)}{R}\)
2 \(\dfrac{R+h}{2}\)
3 \(\dfrac{R+h}{R}\)
4 \(\dfrac{R}{R+h}\)
MHTCET - 2020
For More Updates join with Us
PHXI08:GRAVITATION

359697 A satellite is orbiting close to the earth and has a kinetic energy \(K\). The minimum extra kinetic energy required by it to just overcome the gravitation pull of the earth is

1 \(\sqrt 3 \;K\)
2 \(K\)
3 \(2\sqrt 2 \;K\)
4 \(2K\)
KCET - 2019
PHXI08:GRAVITATION

359698 A satellite orbits the earth at a height of 400 \(km\) above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite \(=200 kg\), mass of the earth \( = 6.0 \times {10^{24}}\;kg\), radius of the earth \( = 6.4 \times {10^6}\;m\),
\(G = 6.67 \times {10^{ - 11}}\;N - {m^2}/k{g^2}\)

1 \(5.2 \times {10^{10}}\;J\)
2 \(3 \times {10^6}\;J\)
3 \(4 \times {10^6}\;J\)
4 \(6 \times {10^9}\;J\)
NCERT Exemplar
PHXI08:GRAVITATION

359699 Assertion :
The binding energy of a satellite does not depend upon the mass of the satellite.
Reason :
Binding energy is the negative value of total energy 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.
PHXI08:GRAVITATION

359700 The binding energy of a satellite of mass \(m\) in an orbit of radius \(r\) is (\(R=\) radius of earth, \(g=\) acceleration due to gravity):

1 \(\dfrac{m g R^{2}}{2 r}\)
2 \(\dfrac{m g R^{2}}{r}\)
3 \(-\dfrac{m g R^{2}}{2 r}\)
4 \(-\dfrac{m g R^{2}}{r}\)
PHXI08:GRAVITATION

359701 The ratio of binding energy of a statellite at rest on earth's surface to the binding energy of a statellte of same mass revolving around the earth at a height \(h\) above the earth's surface is (\(R=\) radius of earth)

1 \(\dfrac{2(R+h)}{R}\)
2 \(\dfrac{R+h}{2}\)
3 \(\dfrac{R+h}{R}\)
4 \(\dfrac{R}{R+h}\)
MHTCET - 2020
NEET DIGITAL LIBRARY
PHXI08:GRAVITATION

359697 A satellite is orbiting close to the earth and has a kinetic energy \(K\). The minimum extra kinetic energy required by it to just overcome the gravitation pull of the earth is

1 \(\sqrt 3 \;K\)
2 \(K\)
3 \(2\sqrt 2 \;K\)
4 \(2K\)
KCET - 2019
PHXI08:GRAVITATION

359698 A satellite orbits the earth at a height of 400 \(km\) above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite \(=200 kg\), mass of the earth \( = 6.0 \times {10^{24}}\;kg\), radius of the earth \( = 6.4 \times {10^6}\;m\),
\(G = 6.67 \times {10^{ - 11}}\;N - {m^2}/k{g^2}\)

1 \(5.2 \times {10^{10}}\;J\)
2 \(3 \times {10^6}\;J\)
3 \(4 \times {10^6}\;J\)
4 \(6 \times {10^9}\;J\)
NCERT Exemplar
PHXI08:GRAVITATION

359699 Assertion :
The binding energy of a satellite does not depend upon the mass of the satellite.
Reason :
Binding energy is the negative value of total energy 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.
PHXI08:GRAVITATION

359700 The binding energy of a satellite of mass \(m\) in an orbit of radius \(r\) is (\(R=\) radius of earth, \(g=\) acceleration due to gravity):

1 \(\dfrac{m g R^{2}}{2 r}\)
2 \(\dfrac{m g R^{2}}{r}\)
3 \(-\dfrac{m g R^{2}}{2 r}\)
4 \(-\dfrac{m g R^{2}}{r}\)
PHXI08:GRAVITATION

359701 The ratio of binding energy of a statellite at rest on earth's surface to the binding energy of a statellte of same mass revolving around the earth at a height \(h\) above the earth's surface is (\(R=\) radius of earth)

1 \(\dfrac{2(R+h)}{R}\)
2 \(\dfrac{R+h}{2}\)
3 \(\dfrac{R+h}{R}\)
4 \(\dfrac{R}{R+h}\)
MHTCET - 2020