If the gravitational field in the space is given as (-(K/r2)). Taking the reference point to be at r =2 cm with gravitational potential V =10 J / kg. Find the gravitational potential at r =3 cm in SI unit (Given, that K =6 Jcm / kg )

Question

If the gravitational field in the space is given as (-(K/r2)). Taking the reference point to be at r =2 cm with gravitational potential V =10 J / kg. Find the gravitational potential at r =3 cm in SI unit (Given, that K =6 Jcm / kg ) (A) 11 (B) 10 (C) 12 (D) 9

Tardigrade Admin · Accepted Answer

-( dV / dr )=-( k / r 2) ⇒ ∫ limits10 V dV =∫ limits23 ( k / r 2) dr V -10= k [(1/2)-(1/3)] V -10=( k /6) ⇒ V =11 text volts

Q. If the gravitational field in the space is given as $(- \frac{K}{r ^{2}})$ . Taking the reference point to be at $r = 2 c m$ with gravitational potential $V = 10 J / k g$ . Find the gravitational potential at $r = 3 c m$ in SI unit (Given, that $K = 6 J c m / k g$ )

11

10

12

9

$- \frac{d V}{d r} = - \frac{k}{r ^{2}} \Rightarrow 10 \int V d V = 2 \int 3 \frac{k}{r ^{2}} d r$
$V - 10 = k [\frac{1}{2} - \frac{1}{3}]$
$V - 10 = \frac{k}{6} \Rightarrow V = 11 volts$

Q. If the gravitational field in the space is given as (−r2K​). Taking the reference point to be at r=2cm with gravitational potential V=10J/kg. Find the gravitational potential at r=3cm in SI unit (Given, that K=6Jcm/kg )

11

10

12

9

−drdV​=−r2k​⇒10∫V​dV=2∫3​r2k​dr V−10=k[21​−31​] V−10=6k​⇒V=11 volts

Q. If the gravitational field in the space is given as $(- \frac{K}{r ^{2}})$ . Taking the reference point to be at $r = 2 c m$ with gravitational potential $V = 10 J / k g$ . Find the gravitational potential at $r = 3 c m$ in SI unit (Given, that $K = 6 J c m / k g$ )

$- \frac{d V}{d r} = - \frac{k}{r ^{2}} \Rightarrow 10 \int V d V = 2 \int 3 \frac{k}{r ^{2}} d r$
$V - 10 = k [\frac{1}{2} - \frac{1}{3}]$
$V - 10 = \frac{k}{6} \Rightarrow V = 11 volts$