If a stone is thrown by hand, we see it falls back to the earth. Of course using machines we can shoot an object with much greater speeds and with greater and greater initial speed, the object scales higher and higher heights. A natural query that arises in our mind is the following: ‘can we throw an object with such high initial speeds that it does not fall back to the earth?’
The principle of conservation of energy helps us to answer this question. Suppose the object did reach infinity and that its speed there was $V_f$ . The energy of an object is the sum of potential and kinetic energy. As before $W_1$ denotes that gravitational potential energy of the object at infinity. The total energy of the projectile at infinity then is
$E(\infty)=W_1+\frac{mV_f^2}{2}$ (1)
If the object was thrown initially with a speed $V_i$ from a point at a distance ($h+R_E$) from the centre of the earth ($R_E$ = radius of the earth), its energy initially was
$E(h+R_E)=\frac{1}{2}mV_i^2-\frac{GmM_E}{(h+R_E)}+W_1$ (2)
By the principle of energy conservation Eqs. (8.26) and (8.27) must be equal. Hence
$\frac{1}{2}mV_i^2-\frac{GmM_E}{(h+R_E)}=\frac{1}{2}mV_f^2$ (3)
The R.H.S. is a positive quantity with a minimum value zero hence so must be the L.H.S. Thus, an object can reach infinity as long as $V_i$ is such that
$\frac{1}{2}mV_i^2-\frac{GmM_E}{(h+R_E)} \geqslant 0$ (4)
The minimum value of Vi corresponds to the case when the L.H.S. of Eq. (4) equals zero.
Thus, the minimum speed required for an object to reach infinity (i.e. escape from the earth) corresponds to
$\frac{1}{2}m(V_i)_{min}^2=\frac{GmM_E}{(h+R_E)}$ (5)
If the object is thrown from the surface of the earth, h = 0, and we get
$(V_i)_{min}=\sqrt{\frac{GM_E}{h+R_E}}$ (6)
Using the relation $g=GM_E/R_E^2$
$(V_i)_{min}=\sqrt{2gR_E}$ (7)
Using the value of g and RE, numerically
$(V_i)_{min}$ ≈ 11.2 km/s.
This is called the escape speed, sometimes loosely called the escape velocity.
Equation (7) applies equally well to an object thrown from the surface of the moon with g replaced by the acceleration due to Moon’s gravity on its surface and $r_E$ replaced by the radius of the moon. Both are smaller than their values on earth and the escape speed for the moon turns out to be 2.3 km/s, about five times smaller. This is the reason that moon has no atmosphere. Gas molecules if formed on the surface of the moon having velocities larger than this will escape the gravitational pull of the moon.
Example 1
Two uniform solid spheres of equal radii R, but mass M and 4 M have a centre to centre separation 6 R, as shown in Fig. 1. The two spheres are held fixed. A projectile of mass m is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere.
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| Fig.1 |
The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point N (see Fig. 1 is defined as the position where the two forces cancel each other exactly. If ON = r, we have
$\frac{GMm}{r^2}=\frac{4GMm}{(6R-r)^2}$
$(6R-r)^2=4r^2$
6R – r = ±2r
r = 2R or –6R.
The neutral point r = –6R does not concern us in this example. Thus ON = r = 2R. It is sufficient to project the particle with a speed which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface of M is
$E_i=\frac{1}{2}mv^2-\frac{GMm}{R}-\frac{4GMm}{5R}$
At the neutral point N, the speed approaches zero. The mechanical energy at N is purely potential.
$E_N=-\frac{GMm}{2R}-\frac{4GMm}{4R}$
From the principle of conservation of mechanical energy
$\frac{1}{2}v^2-\frac{GM}{R}-\frac{4GM}{5R}=-\frac{GM}{2R}-\frac{GM}{R}$
or
$v^2=\frac{2GM}{R}\left(\frac{4}{5}-\frac{1}{2}\right)$
$v=\sqrt{\frac{3GM}{5R}}$
A point to note is that the speed of the projectile is zero at N, but is nonzero when it strikes the heavier sphere 4M. The calculation of this speed is left as an exercise to the students.
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