2. When a body is projected vertically
upwards from the surface of the earth with
certain velocity, it will attain certain height and
return to the ground due to gravitational force.
As the velocity is increased, it will attain
a greater height before falling. If the velocity is so
large that it crosses the gravitational field, then
we can say that body is escaped.
2
3. The escape velocity on the earth
is the minimum velocity with which the
body has to be projected vertically
upwards from the surface of the earth so
that it crosses the gravitational field of
earth.
3
4. Suppose M and R is the mass
and radius of the earth with center
at O , as show as the figure.
Suppose a body of mass m
Is to be projected from the surface
of earth. Consider two points P and
Q at a distance x and x + dx from the
center of earth
4
5. Gravitational force on the body at
point P is F =
𝐺𝑀𝑚
𝑥2
Now, small amount of work dw
has to be done on the body against this
gravitational force in taking the body
from P to Q.
Therefore, dw = F dx =
𝐺𝑀𝑚
𝑥2 dx
5
6. To calculate total work done on
the body in taking it against gravitational
force from surface of earth to a region
beyond the gravitational field of earth,
Integrate the above equation between the
limits x – R to x = ∞.
6
7. This work is done on the body at the cost
of kinetic energy, given to the body at the
surface of earth. If ve is the escape
velocity, then kinetic energy of the body
will be
1
2
mve
2 .
Therefore,
1
2
mve
2 =
𝐺𝑀𝑚
𝑅
.
Hence, ve =
2𝐺𝑀
𝑅
7
8. But g =
𝐺𝑀
𝑅2 .
Therefore , GM = gR2. Using this value in the
above equation,
We get ve = 2𝑔R .
As we know, g = 9.8 m/s2 and
radius of earth = 6.4 × 106 m,
We get ve = 11.2 km/s .
8
