An object weights 72 N on earth. Its weight at a height of R /2 from earth is

Options

A. `32 N`
B. `56 N`
C. `72 N`
D. Zero

Answer

To find the weight of an object at a height of R/2 from the surface of the Earth, we can follow these steps:

### Step 1: Understand the Weight of the Object
The weight of the object on the surface of the Earth is given as 72 N. This weight can be expressed using the formula:
\[ W = \frac{GMm}{R^2} \]
where:
- \( W \) is the weight of the object,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the object,
- \( R \) is the radius of the Earth.

### Step 2: Determine the Height Above the Earth's Surface
The height from the Earth's surface is given as \( R/2 \). Therefore, the distance from the center of the Earth to the object at this height is:
\[ \text{Distance from center} = R + \frac{R}{2} = \frac{3R}{2} \]

### Step 3: Calculate the Weight at the New Height
The weight of the object at this new height can be calculated using the formula:
\[ W' = \frac{GMm}{(3R/2)^2} \]
This simplifies to:
\[ W' = \frac{GMm}{\frac{9R^2}{4}} = \frac{4GMm}{9R^2} \]

### Step 4: Relate the New Weight to the Original Weight
We know from the original weight on the surface:
\[ W = \frac{GMm}{R^2} = 72 \, \text{N} \]
Now we can express the new weight \( W' \) in terms of the original weight:
\[ W' = \frac{4}{9} W \]
Substituting the value of \( W \):
\[ W' = \frac{4}{9} \times 72 \]

### Step 5: Perform the Calculation
Calculating \( W' \):
\[ W' = \frac{4 \times 72}{9} = \frac{288}{9} = 32 \, \text{N} \]

### Final Answer
The weight of the object at a height of \( R/2 \) from the Earth's surface is **32 N**.

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