Question

Calculate the escape velocity of a body from the surface of the earth.
[Given: $G=6.67 × 10^{-11}$ $Nm^2/{kg}^2$, $R= 6400 km$, $g= 9.8 m/s^2$]

Answer 1

Hint: The given problem is related to gravitation. The escape velocity of any body is proportional to square root of mass of body and inversely proportional to square root of radius of planet. So we use this relation to calculate escape velocity of a body from the surface of earth.

Complete step by step solution:
As we know that the escape velocity is defined as the minimum velocity of an object, due to which it crosses the gravitational field of the plant or is not attracted by the planet. The escape velocity of object from any planet (here earth) will be given by:
${v_e} = \sqrt {\dfrac{{2GM}}{R}} $
Here, M= Mass of Planet
R= Radius of planet
G= Gravitational constant
The values of these parameters are as following
R= 6400 km
$G=6.67×10^{-11} Nm^2/{kg}^2$
And the mass of earth is, $M= 6×10^{24} kg$
After putting the values of these parameters in the formula of escape velocity, we will find the value of escape velocity of the body for earth.
${v_e} = \sqrt {\dfrac{{2 \times 6.67 \times {{10}^{ - 11}} \times 6 \times {{10}^{24}}}}{{6400}}} \\
{v_e} = 11.2km/\sec \\ $
So the value of escape velocity of any object from earth will be 11.2 km/sec. If we will project any object with this velocity from earth's surface then it crosses the gravitational effect of earth and never comes back on earth’s surface.

Note: The escape velocity depends only on the mass and size of the object from which something is trying to escape. The concept of escape velocity is used to determine the factors that are important to project a satellite or missiles from one planet to another planet.