Question

The masses and radii of the earth and moon are m1, r1 and m2, r2 respectively. Their centres are at distance d apart. The minimum speed with which a particle of mass m should be projected from a point midway the two centres so as to escape to infinity is

Answer 1

from the law of conservation of energy,
∆K.E = - ∆P.E .......(1)

here, change in kinetic energy, ∆K.E = 1/2 mv², where v is the minimum speed of particle of mass m should be projected from a point midway the two centres so as to escape to infinity.

change in potential energy, ∆P.E = -Gm1m/(d/2) - Gm2m/(d/2) - 0

= -2Gm/d [ m1 + m2]

now from equation (1),

1/2 mv² = 2Gm/d (m1 + m2)

v² = 4G/d (m1 + m2)

v = $\sqrt{\frac{4G}{d}(m_1+m_2)}$

Answer 2

Answer:

Explanation:

from the law of conservation of energy,

∆K.E = - ∆P.E .......(1)

here, change in kinetic energy, ∆K.E = 1/2 mv², where v is the minimum speed of particle of mass m should be projected from a point midway the two centres so as to escape to infinity.

change in potential energy, ∆P.E = -Gm1m/(d/2) - Gm2m/(d/2) - 0

= -2Gm/d [ m1 + m2]

now from equation (1),

1/2 mv² = 2Gm/d (m1 + m2)

v² = 4G/d (m1 + m2)

v = $\sqrt{(4G(m1+m2))/d$