If the gravitational force had varied as `r^(-5//2)` instead of `r^(-2)` the potential energy of a particle at a distance 'r' from the centre of the earth would be proportional to

Options

A. `r^(-1)`
B. `r^(-2)`
C. `r^(-3//2)`
D. `r^(-5//2)`

Answer

To find the potential energy of a particle at a distance 'r' from the center of the Earth when the gravitational force varies as \( r^{-5/2} \), we can follow these steps:

### Step 1: Understand the relationship between force and potential energy
The potential energy \( U(r) \) for a conservative force can be found by integrating the force from infinity to a distance \( r \):
\[ U(r) = -\int_{\infty}^{r} F \, dr \]

### Step 2: Define the gravitational force
The gravitational force \( F \) is given as:
\[ F = \frac{G m_1 m_2}{r^{5/2}} \]
where \( G \) is the gravitational constant, and \( m_1 \) and \( m_2 \) are the masses involved.

### Step 3: Set up the integral for potential energy
Substituting the expression for \( F \) into the potential energy equation, we have:
\[ U(r) = -\int_{\infty}^{r} \frac{G m_1 m_2}{r^{5/2}} \, dr \]

### Step 4: Perform the integration
To integrate \( \frac{1}{r^{5/2}} \), we can rewrite it as:
\[ U(r) = -G m_1 m_2 \int_{\infty}^{r} r^{-5/2} \, dr \]
The integral of \( r^{-5/2} \) is:
\[ \int r^{-5/2} \, dr = -\frac{2}{3} r^{-3/2} \]
Thus, we have:
\[ U(r) = -G m_1 m_2 \left[ -\frac{2}{3} r^{-3/2} \right]_{\infty}^{r} \]

### Step 5: Evaluate the limits of the integral
Evaluating the limits, we find:
\[ U(r) = -G m_1 m_2 \left( -\frac{2}{3} r^{-3/2} - 0 \right) = \frac{2}{3} \frac{G m_1 m_2}{r^{3/2}} \]

### Step 6: Write the final expression
Thus, the potential energy \( U(r) \) is proportional to:
\[ U(r) \propto \frac{1}{r^{3/2}} \]

### Conclusion
The potential energy of a particle at a distance \( r \) from the center of the Earth, when the gravitational force varies as \( r^{-5/2} \), is proportional to \( r^{-3/2} \).

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