Let the gravitational field intensity due to the point masses m₁ and m₂ be zero at P

⇒  g_p = 0    ⇒ |g₁ + g₂| = 0

⇒ g₁ = g₂ (numerically)

⇒ \displaystyle \frac{{G{{m}_{1}}}}{{{{x}^{2}}}}= \displaystyle \frac{{G{{m}_{2}}}}{{{{{(\ell -x)}}^{2}}}}

⇒ \displaystyle \frac{1}{{{{x}^{2}}}}= \displaystyle \frac{4}{{{{{(3-x)}}^{2}}}}

⇒  \displaystyle \frac{{3-x}}{x}= ±2

⇒ 3 – x = – 2x or,          3 – x = +2x

⇒ x = -3 (irrelevant) or x = 1

V_p = \displaystyle \left( {-\frac{{G{{m}_{1}}}}{x}} \right) + \displaystyle \left( {-\frac{{G{{m}_{2}}}}{{\ell -x}}} \right)

=  – G \displaystyle \left[ {\frac{{{{m}_{1}}}}{x}+\frac{{{{m}_{2}}}}{{\ell -x}}} \right]

 Putting G = 6.67 ´ 10^-11 N-m²/kg²

 m₁ = 1 kg, m₂ = 4 kg, l = 3 m, x = 1 m, we obtain,

 V_p = -6.67 x 10^–11 \displaystyle \left( {\frac{1}{1}+\frac{4}{{3-1}}} \right) J/kg

= -20.01 x 10^-11 J/kg.

Escape Velocity

It is the minimum speed with which a body must be projected from the surface of a planet (usually the earth) so that it permanently overcomes and escapes from the gravitational field of the planet (the earth). We can also say that a body projected with escape speed will be able to go to a point which is at infinite distance from the earth.

If a body of mass m is projected with speed v from the surface of a planet of mass M and radius R, then

K.E. = \frac{1}{2}m{{v}^{2}}\ ;\ \ G.P.E.=-\frac{{GMm}}{R}

Total mechanical energy (T.M.E.) of the body = \frac{1}{2}m{{v}^{2}}-\frac{{GMm}}{R}

If the v’ is the speed of the body at infinity, then

T.M.E. at infinity = 0+\frac{1}{2}mv{{‘}^{2}}=\frac{1}{2}mv{{‘}^{2}}

Applying the principle of conservation of mechanical energy, we have

\frac{1}{2}m{{v}^{2}}-\frac{{GMm}}{R}=\frac{1}{2}mv{{‘}^{2}}

or,          {{v}^{2}}=\frac{{2GM}}{R}+v{{‘}^{2}}

v will be minimum when v’→ o, i.e.,

{{V}_{e}}={{V}_{{\min }}}=\sqrt{{\frac{{2GM}}{R}}} =\sqrt{{2gR}}\             \ \ \ \ \ \left[ {\because \ g=\frac{{GM}}{{{{R}^{2}}}}} \right]

Important points

(i)            Escape speed is independent of the mass and direction of projection of the body.

(ii)           For earth as g = 9.8 m/s² and R = 6400 km

{{V}_{e}}=\sqrt{{2\times 9.8\times 6.4\times {{{10}}^{6}}}}=11.2\ \text{km/s}

Illustration

The masses and radii of the earth and the moon are M₁, R₁ and M₂, R₂ respectively. Their centers are at a distance d apart. What is the minimum speed with which a particle of mass m should be projected from a point midway between the two centers so as to escape to infinity?

Solution:            

Potential energy of m which is midway between M₁ and M₂ is

\displaystyle U=m({{V}_{1}}+{{V}_{2}})=m(-\frac{{G{{m}_{1}}}}{{d/2}}+\frac{{-Gm2}}{{d/2}})

=-\frac{{2Gm}}{d}\ ({{M}_{1}}+{{M}_{2}})

Let v be the required speed, then

(T.M.E.)_initial = -\frac{{2Gm}}{d}\ \left( {{{M}_{1}}+{{M}_{2}}} \right)+\frac{1}{2}m{{v}^{2}}

As the particle reaches infinity,

(T.M.E.)_final = 0

From the principle of conservation of mechanical energy, we have

-\frac{{2Gm}}{d}\,({{M}_{1}}+{{M}_{2}})+\frac{1}{2}m{{v}^{2}}=0

v =2\sqrt{{\frac{{G\ ({{M}_{1}}+{{M}_{2}})}}{d}}}]

Illustration

What will be the escape speed on the surface of the moon if its radius were \frac{1}{4}th the radius of the earth and its mass \left( {\frac{1}{{80}}} \right)th the mass of the earth. Take escape velocity as 11.2 km/s on the surface of the earth. Given g_(earth) = 9.8 m/s².

Solution:            

As on the surface of moon     {{V}_{M}}=\sqrt{{\frac{{2G{{M}_{M}}}}{{{{R}_{M}}}}}}

Further, on earth the  escape speed is     {{V}_{E}}=\sqrt{{\frac{{2G{{M}_{E}}}}{{{{R}_{E}}}}}},

on dividing them

\frac{{{{V}_{M}}}}{{{{V}_{E}}}}=\sqrt{{\frac{{{{M}_{M}}}}{{{{M}_{E}}}}\times \frac{{{{R}_{E}}}}{{{{R}_{M}}}}}}=\sqrt{{\frac{1}{{80}}\times 4}}=\frac{1}{{\sqrt{{20}}}}

{{V}_{M}}=\frac{{{{V}_{E}}}}{{\sqrt{{20}}}}=\frac{{11.2}}{{4.47}} =2.5\ \text{km/s}