Question

If two masses m1 and m2 are present at 'd' distance from each other than prove that potential at which point intensity is 0,
V = -G/d ( m1 + m2 + 2√m1√m2 )​

Answer 1

Explanation:

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Answer 2

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♡ The gravitational force of attraction on m

1 due to m 2

at a separation r is

F

1

=

r

2

Gm

1

m

2

Therefore, the acceleration of m

1

is

a

1

=

m

1

F

1

=

r

2

Gm

2

Similarly, the acceleration of m

2

due to m

1

is

a

2

=−

r

2

Gm

1

The negative sign is put as a

2

is directed opposite to a

1

. The relative acceleration of approach is

a=a

1

−a

2

=

r

2

G(m

1

+m

2

)

(i)

If v is the relative velocity, then

a=

dt

dv

=

dr

dv

dt

dr

But −dr/dt=v(negative sign shows that r decreases with increasing t).

∴a=−

dr

dv

v (ii)

vdv=−

r

2

G(m

1

+m

2

)

dr

Integrating, we get

2v 2

=

r

G(m 1 +m 2 ) +C

At r=∞,v=0(given), and so C=0. Therefore,

v 2 = r

2G(m 1 +m 2 )

Let v=v

R

when r=R. Then

v R

=

R

2(Gm

1 +m 2 )

♡

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