If the gravitational field intensity a point is given by displaystyle g-frac-GM-r-2-5-Then the potential distance r isdisplaystyle frac-2GM-3r-1-5-displaystyle frac-GM-r-3-5-displaystyle frac-GM-3r-1-5-displaystyle g-frac-GM-r-3-5-

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Toppr Admin · Accepted Answer

V-x222B-EdrV-x222B-rinfinityGMr2-5drV-2GM3-x2212-1r1-5-rinfinityV-x2212-2GM3r1-5

If the gravitational field intensity at a point is given by $g = \frac{G M}{r^{2.5}}$ Then the potential at distance $r$ is
$\frac{- 2 G M}{3 r^{1.5}}$
$\frac{- G M}{r^{3.5}}$
$\frac{G M}{3 r^{1.5}}$
$g = \frac{G M}{r^{3.5}}$

$V = \int E d r$

$V = \int_{i n f i n i t y}^{r} \frac{G M}{r^{2.5}} d r$

$V = \frac{2 G M}{3} [- \frac{1}{r^{1.5}}]_{i n f i n i t y}^{r}$

$V = - \frac{2 G M}{3 r^{1.5}}$

If the gravitational field intensity at a point is given by g=GMr2.5Then the potential at distance r is−2GM3r1.5−GMr3.5GM3r1.5g=GMr3.5

V=∫EdrV=∫rinfinityGMr2.5drV=2GM3[−1r1.5]rinfinityV=−2GM3r1.5

If the gravitational field intensity at a point is given by $g = \frac{G M}{r^{2.5}}$ Then the potential at distance $r$ is
$\frac{- 2 G M}{3 r^{1.5}}$
$\frac{- G M}{r^{3.5}}$
$\frac{G M}{3 r^{1.5}}$
$g = \frac{G M}{r^{3.5}}$

$V = \int E d r$

$V = \int_{i n f i n i t y}^{r} \frac{G M}{r^{2.5}} d r$

$V = \frac{2 G M}{3} [- \frac{1}{r^{1.5}}]_{i n f i n i t y}^{r}$

$V = - \frac{2 G M}{3 r^{1.5}}$