Problem 59 - Master Resource Book in JEE Main Physics

Question

In a certain region of space, the gravitational field is given by \(-k / r\), where \(r\) is the distance and \(k\) is a constant. If the gravitational potential at \(r=r_{0}\) be \(V_{0}\), then what is the expression for the gravitational potential \(V\) ? (a) \(k \log \left(\frac{r}{r_{0}}\right)\) (b) \(k \log \left(\frac{r_{0}}{r}\right)\) (c) \(V_{0}+k \log \left(\frac{r}{r_{0}}\right)\) (d) \(V_{0}+k \log \left(\frac{r_{0}}{r}\right)\)

Step-by-Step Solution

Answer

The correct expression for the gravitational potential is option (d): \(V_0 + k \ln\left( \frac{r_0}{r} \right)\).

Step 1: Understand the Gravitational Field and Potential Relation

The gravitational field is given by \( -\frac{k}{r} \). The relationship between gravitational field \( E \) and gravitational potential \( V \) is \( \frac{dV}{dr} = -E \), which allows us to find the potential from the field.

Step 2: Set Up the Integral Equation

We know that \( \frac{dV}{dr} = -\frac{k}{r} \). We can find \( V \) by integrating both sides with respect to \( r \): \[ dV = -\frac{k}{r} dr \] \( V = -\int \frac{k}{r} dr \).

Step 3: Perform the Integration

Integrate the right-hand side: \[ V = -k \int \frac{1}{r} dr = -k \ln(r) + C \] where \( C \) is the constant of integration.

Step 4: Apply Boundary Conditions

Given that at \( r = r_0 \), the potential is \( V_0 \), substitute into the equation to find \( C \): \[ V_0 = -k \ln(r_0) + C \]. Solving for \( C \), we get \( C = V_0 + k \ln(r_0) \).

Step 5: Express the Gravitational Potential

Substitute \( C \) back into the expression for \( V \): \[ V = -k \ln(r) + V_0 + k \ln(r_0) \].Simplify:\[ V = V_0 + k \ln\left( \frac{r_0}{r} \right) \].

Step 6: Choose the Correct Option

From the multiple-choice answers provided, the expression we found matches option (d): \[ V_0 + k \ln\left( \frac{r_0}{r} \right) \].

Key Concepts

Gravitational Field

In physics, a gravitational field is a model used to explain the influence that a massive body extends into the space around itself. This allows us to understand how a mass creates a force of attraction. In this exercise, the gravitational field is described by the function \[-\frac{k}{r}\]. Here, \(k\) is a constant, and \(r\) represents the distance from the source of the field.

Gravitational fields usually point towards the mass creating them. In this case, the negative sign indicates that the force is attractive, pulling objects towards the mass. Understanding the gravitational field requires us to relate it to gravitational potential, since fields indicate how the potential changes in space.

Key points to remember:

• Gravitational fields describe how objects are influenced by gravity.
• The formula \(-\frac{k}{r}\) tells us the strength and direction of the field based on distance.
• The negative sign signifies that the force is an attractive one, pointing towards the source.

Integration

Integration is a mathematical process used to find a function that describes accumulated quantities, such as areas under curves or total accumulation of change given a rate. In this context, we use integration to go from the gravitational field to gravitational potential.

The solution involves the integral: \[ V = -k \int \frac{1}{r} \, dr \]
This step transforms the expression of gravitational field into the expression for gravitational potential. When we perform this integral, we are essentially summing up the infinitesimal changes in potential over distance \(r\). It leads us to a logarithmic function in this scenario, as the integral of \(\frac{1}{r}\) is the natural logarithm, \(\ln(r)\).

In summary:

• Integration helps convert rate-based information (field) into accumulated information (potential).
• It involves finding the indefinite integral of the given expression.

Boundary Conditions

Boundary conditions are essential to solving differential equations where constants of integration appear. They serve as specific conditions or constraints that the solutions must meet at certain points.

In this exercise, we know that when \(r = r_0\), the gravitational potential \(V\) is \(V_0\). This information allows us to calculate the constant of integration \(C\). By applying the boundary condition in the equation:\[ V_0 = -k \ln(r_0) + C \],
we solve for \(C\), which integrates our specific condition into the general solution.
Thus, boundary conditions provide:

• Specific values that tailor the general integral solution to match real-world or given parameters.
• Stability in solutions by anchoring them to particular known values.

Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It is an important concept because it represents the "stored" energy that could be converted into kinetic energy as the object moves.

In the context of this problem, gravitational potential \(V\) is calculated from the gravitational field and is closely related to potential energy. The potential energy per unit mass at a point is given by the gravitational potential, so knowing \(V\) helps determine how much work would be needed to move an object from a reference point.
Main takeaways about gravitational potential energy:

• It is directly related to the gravitational potential \(V\).
• Represents the energy stored due to position relative to a source.
• Understanding \(V\) helps gauge how much energy changes as positions change.

Logarithmic Functions

Logarithmic functions are mathematical functions that have a logarithm (\(\log\)) involved. They are inverses of exponential functions and arise naturally in various scientific equations.

In our solution, integration of the field yields a function that includes a logarithm: \(-k \ln(r) + C\). The natural logarithm \(\ln\) shows how the potential changes inversely with distance. The use of a logarithmic function implies that the potential changes quickly at small distances and slowly at larger distances.
Important insights into logarithmic functions in this context:

• Appear in the integration process of expressions with \(\frac{1}{r}\).
• Illustrate situations where quantities decrease as they move away from a central point.
• Help capture relationships where changes are rapid initially but taper off.