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

The Correct Option is B

To find the gravitational potential at a distance \( r = 3 \, \text{cm} \) from a reference point, we need to integrate the given gravitational field expression. Here is the step-by-step solution:

The gravitational field strength is given by:

\(g = -\frac{K}{r^2}\)

where \( K = 6 \, \text{J cm/kg} \).

Gravitational potential \( V \) at a distance \( r \) is given by the integral of gravitational field \( g \):

\(V(r) = -\int g \, dr\)

Substituting the expression for \( g \), we have:

\(V(r) = -\int -\frac{K}{r^2} \, dr = \int \frac{K}{r^2} \, dr\)

When calculating the definite integral from the reference point \( r=2 \, \text{cm} \) to \( r=3 \, \text{cm} \):

\(\Delta V = V(3) - V(2) = \int_{2}^{3} \frac{K}{r^2} \, dr\)

Let's evaluate this integral:

\(\Delta V = \left[-\frac{K}{r}\right]_{2}^{3} = -\frac{K}{3} - \left(-\frac{K}{2}\right)\)

Substitute \( K = 6 \, \text{J cm/kg} \):

\(\Delta V = -\frac{6}{3} + \frac{6}{2} = -2 + 3 = 1 \, \text{J/kg}\)

Given that the potential at \( r = 2 \, \text{cm} \) is \( V = 10 \, \text{J/kg} \), the potential at \( r = 3 \, \text{cm} \) is:

\(V(3) = V(2) + \Delta V = 10 + 1 = 11 \, \text{J/kg}\)

Therefore, according to the options given, the gravitational potential at \( r = 3 \, \text{cm} \) is 11 J/kg. The correct answer should be adjusted from the given answer.