Question

If the gravitational field in the space is given as $\left(-\frac{K}{r^2}\right)$. 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

Answer 1

Answer: (A)

$ -\frac{ dV }{ dr }=-\frac{ k }{ r ^2} \Rightarrow \int\limits_{10}^{ V } dV =\int\limits_2^3 \frac{ k }{ r ^2} dr $
$ V -10= k \left[\frac{1}{2}-\frac{1}{3}\right] $
$V -10=\frac{ k }{6} \Rightarrow V =11 \text { volts }$