How much energy is necessary for a body of $500 \, kg$ to escape from the Earth? $[g = 9.8 \, m/s^2$,radius of Earth $R = 6.4 \times 10^6 \, m]$

  • A
    About $9.8 \times 10^6 \, J$
  • B
    About $6.4 \times 10^8 \, J$
  • C
    About $3.1 \times 10^{10} \, J$
  • D
    About $27.4 \times 10^{12} \, J$

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Similar Questions

Given below are two statements:
Statement $I:$ For a planet,if the ratio of mass of the planet to its radius increases,the escape velocity from the planet also increases.
Statement $II:$ Escape velocity is independent of the radius of the planet.
In the light of the above statements,choose the most appropriate answer from the options given below:

To project a body of mass $m$ from Earth's surface to infinity,the required kinetic energy is (assume,the radius of Earth is $R_E$,$g =$ acceleration due to gravity on the surface of Earth):

Escape velocity at the surface of the Earth is $11.2 \, km/s$. If the radius of a planet is double that of the Earth but its mean density is the same as that of the Earth,then the escape velocity will be ........ $km/s$.

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If a planet of mass $6.4 \times 10^{23} \ kg$ can be compressed into a sphere such that the escape velocity from its surface is $8 \times 10^4 \ m/s$,then what should be the radius of the sphere (in $km$)? (Gravitational constant,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$)

$A$ body is projected vertically upwards from the surface of the earth with a velocity sufficient to carry it to infinity. The time taken by it to reach a height of three times the radius of the earth is (acceleration due to gravity $g = 9.8 \ m/s^2$ and radius of the earth $R = 6400 \ km$). (in $min$)

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