A mass $M$ is split into two parts $m$ and $(M - m)$ , which are then separated by a certain distance. What ratio $(m / M)$ maximizes the gravitational force between the parts.

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Solution

$F = \frac{G m (M - m)}{R^{2}} = \frac{g (m M - m^{2})}{R^{2}}$

Differentiating w.r.t m and equating to $0$ for maximizing the force-

$⟹ \frac{d F}{d m} = \frac{G}{R^{2}} (M - 2 m) = 0$

$⟹ M = 2 m$

$⟹ \frac{m}{M} = \frac{1}{2}$

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A mass M is split into two parts, m and (M-m), which are then separated by a certain distance. What ratio of $\frac{m}{M}$ maximizes the gravitational force between the two parts

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Q. A mass of

M

at rest is broken into two pieces having masses

m

and

(M - m)

. The two masses are then separated by a distance. The gravitational force between them will be the maximum when the ratio of the masses [

m : (M - m)

] of the two parts is:

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A mass M is split into two parts m and (M−m), which are then separated by a certain distance. What ratio (m/M) maximizes the gravitational force between the parts.

F=Gm(M−m)R2=g(mM−m2)R2Differentiating w.r.t m and equating to 0 for maximizing the force-⟹dFdm=GR2(M−2m)=0⟹M=2m⟹mM=12

A mass $M$ is split into two parts $m$ and $(M - m)$ , which are then separated by a certain distance. What ratio $(m / M)$ maximizes the gravitational force between the parts.

$F = \frac{G m (M - m)}{R^{2}} = \frac{g (m M - m^{2})}{R^{2}}$

Differentiating w.r.t m and equating to $0$ for maximizing the force-

$⟹ \frac{d F}{d m} = \frac{G}{R^{2}} (M - 2 m) = 0$

$⟹ M = 2 m$

$⟹ \frac{m}{M} = \frac{1}{2}$