Page 207 - Class-11-Physics-Part-1_Neat

GRAVITATION 193

Thus, the minimum speed required for an object
2 G m 2  1  G m 2
= −   2 +   = − 5.41 to reach infinity (i.e. escape from the earth)
l  2  l corresponds to
The gravitational potential at the centre of 1 2 GmM
( )
m V i = E (8.30)
the square (r = 2 l / ) 2 is 2 min h + R E
If the object is thrown from the surface of
G m
)
U( r = − 4 2 the earth, h = 0, and we get
l . t
2GM
V
8.8 ESCAPE SPEED ( ) min = E (8.31)
i
R E
If a stone is thrown by hand, we see it falls back
2
to the earth. Of course using machines we can Using the relation g = GM E /R , we get
E
shoot an object with much greater speeds and
with greater and greater initial speed, the object ( ) = 2gR E (8.32)
V
i
scales higher and higher heights. A natural min
query that arises in our mind is the following: Using the value of g and R , numerically
E
‘can we throw an object with such high initial (V ) i min ≈11.2 km/s. This is called the escape
speeds that it does not fall back to the earth?’ speed, sometimes loosely called the escape
The principle of conservation of energy helps velocity.
us to answer this question. Suppose the object Equation (8.32) applies equally well to an
did reach infinity and that its speed there was object thrown from the surface of the moon with
V . The energy of an object is the sum of potential g replaced by the acceleration due to Moon’s
f gravity on its surface and r replaced by the
and kinetic energy. As before W denotes that E
1 radius of the moon. Both are smaller than their
gravitational potential energy of the object at
infinity. The total energy of the projectile at values on earth and the escape speed for the
infinity then is 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
2
mV
E ( ) W∞ = 1 + f (8.26) surface of the moon having velocities larger than
2 this will escape the gravitational pull of the
If the object was thrown initially with a speed moon.
V from a point at a distance (h+R ) from the
i E t
centre of the earth (R = radius of the earth), its Example 8.4 Two uniform solid spheres
E
energy initially was of equal radii R, but mass M and 4 M have
a centre to centre separation 6 R, as shown
1 GmM E
E (h + R E ) = mV i 2 – + W 1 (8.27) in Fig. 8.10. The two spheres are held fixed.
2 (h + R )
E A projectile of mass m is projected from
By the principle of energy conservation the surface of the sphere of mass M directly
Eqs. (8.26) and (8.27) must be equal. Hence towards the centre of the second sphere.
Obtain an expression for the minimum
2
mV 2 GmM mV f speed v of the projectile so that it reaches
i
– E = (8.28)
2 (h + R ) 2 the surface of the second sphere.
E
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
mV i 2 GmM E
– ≥ 0 (8.29)
2 (h + R ) Fig. 8.10
E
The minimum value of V corresponds to the Answer The projectile is acted upon by two
i
case when the L.H.S. of Eq. (8.29) equals zero. mutually opposing gravitational forces of the two

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