`int_r^infty (Gm_1m_2)/(r^2) dr` is equal to

Options

A. `(Gm_1m_2)/r`
B. `-(Gm_1m_2)/(r^3)`
C. `(Gm_1m_2)/(r^3)`
D. `-(Gm_1m_2)/r`

Answer

To solve the integral

\[ \int_r^{\infty} \frac{G m_1 m_2}{r^2} \, dr \]

we will follow these steps:

### Step 1: Identify the constants
The expression \(\frac{G m_1 m_2}{r^2}\) represents the gravitational force between two masses \(m_1\) and \(m_2\) separated by a distance \(r\). Here, \(G\) is the gravitational constant. Since \(G\), \(m_1\), and \(m_2\) are constants with respect to the variable of integration \(r\), we can factor them out of the integral.

### Step 2: Factor out the constants
We can rewrite the integral as:

\[ G m_1 m_2 \int_r^{\infty} \frac{1}{r^2} \, dr \]

### Step 3: Integrate the function
Now we need to integrate \(\frac{1}{r^2}\):

\[ \int \frac{1}{r^2} \, dr = -\frac{1}{r} + C \]

### Step 4: Evaluate the definite integral
We will evaluate the definite integral from \(r\) to \(\infty\):

\[ \int_r^{\infty} \frac{1}{r^2} \, dr = \left[-\frac{1}{r}\right]_{r}^{\infty} \]

### Step 5: Substitute the limits
Now we substitute the limits into the expression:

\[ = \left(-\frac{1}{\infty}\right) - \left(-\frac{1}{r}\right) \]

### Step 6: Simplify the expression
As \(r\) approaches infinity, \(-\frac{1}{\infty} = 0\):

\[ = 0 + \frac{1}{r} = \frac{1}{r} \]

### Step 7: Combine with the constants
Now, substituting back into our expression with the constants factored out:

\[ G m_1 m_2 \cdot \frac{1}{r} = \frac{G m_1 m_2}{r} \]

### Final Answer
Thus, the value of the integral

\[ \int_r^{\infty} \frac{G m_1 m_2}{r^2} \, dr = \frac{G m_1 m_2}{r} \]