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The gravitati...

Question

The gravitational force of attraction between a uniform sphere of mass $M$ and a uniform rod of length $l$ and mass $m$ oriented as shown is

\frac{G M m}{r (r + l)}

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\frac{G M}{r^{2}}

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M m r^{2} + l

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(r^{2} + l) m M

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Solution

The correct option is A $\frac{G M m}{r (r + l)}$
The force of attraction between sphere and shaded position $d F = G M \frac{(\frac{m}{l} d x)}{x^{2}}$
$F = \int_{r}^{r + 1} \frac{G M m}{l x^{2}} d x = \frac{G M m}{l} \int_{r}^{r + 1} \frac{1}{x^{2}} d x$

$= \frac{G M m}{l} \int_{r}^{r + 1} x^{- 2} d x = \frac{G M m}{l} {[\frac{x^{- 2 + 1}}{- 2 + 1}]}_{r}^{r + 1}$

$= - \frac{G M m}{l} {[x^{- 1}]}_{r}^{- 1} = - \frac{G M m}{l} {[\frac{1}{x}]}_{r}^{r + 1} = \frac{G M m}{r (r + 1)}$

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The gravitational force of attraction between a uniform sphere of mass M and a uniform rod of length l and mass m oriented as shown is

The correct option is A GMmr(r+l)The force of attraction between sphere and shaded position dF=GM(mldx)x2F=∫r+1rGMmlx2dx=GMml∫r+1r1x2dx=GMml∫r+1rx−2dx=GMml[x−2+1−2+1]r+1r=−GMml[x−1]r+1r=−GMml[1x]r+1r=GMmr(r+1)

The gravitational force of attraction between a uniform sphere of mass $M$ and a uniform rod of length $l$ and mass $m$ oriented as shown is