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Q. A mass is split into two parts (m) and (M-m) which are then separated by a certain distance. What ratio M/m will maximise the gravitational force between them?✔
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Heya Guys!!✌
Q. A mass is split into two parts (m) and (M-m) which are then separated by a certain distance. What ratio M/m will maximise the gravitational force between them?✔
❎Don't spam..!!❎
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Answer -
\frac{m}{m } = \frac{1}{2}
m
m
=
2
1
___________________________
Explanation-
if r is the distance between m and (M-m), the gravitational force will be,
f = g \times \frac{m(m - m)}{ {r}^{2} } = \frac{g}{ {r}^{2} } \times (mm - {m}^{2}f=g×
r
2
m(m−m)
=
r
2
g
×(mm−m
2
for f to be maximum,
\frac{df}{dm} = 0 \: and \: \frac{ {d}^{2}f }{d {m}^{2} } < 0
dm
df
=0and
dm
2
d
2
f
<0
as M and r constants,
i.e.
\frac{d}{dm} (mm - {m}^{2} \times \frac{g}{ {r}^{2} } ) = 0
dm
d
(mm−m
2
×
r
2
g
)=0
==>
m - 2m = 0 \: \: \: \: \: \: \: \: \: \: \: \:( \frac{g}{ {r}^{2} } is \:not \: equal \: to \: 0)m−2m=0(
r
2
g
isnotequalto0)
==>
\frac{m}{m} = \frac{1}{2}
m
m
=
2
1
the force will be maximum when the two parta are identical.