A body weighs 64 N on the surface of the Earth. What is the gravitational force on it (in N) due to the Earth at a height equal to one-third of the radius of the Earth?

Explanation

To solve the problem, we need to find the gravitational force acting on a body at a height equal to one-third of the Earth's radius. We know the weight of the body on the surface of the Earth is 64 N.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Weight of the body on the surface of the Earth (W) = 64 N
- Height (h) = \( \frac{1}{3} \) of the Earth's radius (R).

2. **Understand the Relationship Between Weight and Gravitational Force:**
- The weight of the body on the surface of the Earth is given by the formula:
\[ W = mg \]
where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity at the surface of the Earth.

3. **Calculate the Mass of the Body:**
- Rearranging the weight formula gives us:
\[ m = \frac{W}{g} \]
- We will need the value of \( g \) (approximately \( 9.8 \, \text{m/s}^2 \)) to find the mass, but we can also express \( m \) in terms of \( W \) and \( g \).

4. **Determine the New Gravitational Acceleration at Height \( h \):**
- The new height \( h \) is \( \frac{R}{3} \).
- The formula for gravitational acceleration at a height \( h \) above the surface of the Earth is:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
- Substituting \( h = \frac{R}{3} \):
\[ g' = \frac{g}{(1 + \frac{1}{3})^2} = \frac{g}{(\frac{4}{3})^2} = \frac{g}{\frac{16}{9}} = \frac{9g}{16} \]

5. **Calculate the New Weight at Height \( h \):**
- The new weight \( W' \) at height \( h \) can be calculated using:
\[ W' = mg' = m \left(\frac{9g}{16}\right) \]
- Since \( mg = W \), we can substitute:
\[ W' = \frac{9}{16} W \]
- Substituting \( W = 64 \, \text{N} \):
\[ W' = \frac{9}{16} \times 64 = 36 \, \text{N} \]

6. **Final Answer:**
- The gravitational force on the body at a height equal to one-third of the radius of the Earth is **36 N**.