GRAVITATIONAL FORCE.  
Mass of Earth = 5.9742 x 1024 kg	Radius of Earth = 6378 km 	Radius of the Sun = 696,340 km
Mass of Moon = 0.073483 x 1024 kg	Radius of Moon = 1737 km 	
Mass of the Sun = 1.9891 × 1030 kilograms
Distance between Earth and Sun = 149.6 x 106 km      G = 6.67408 x 10-11 m3kg-1s-2                      
 A man weighs 150 N on the surface of the Earth. Find his weight at an elevation equal to the radius of the Earth.

To find the man's weight at an elevation equal to the radius of the Earth, we need to apply the **gravitational force formula**:

\[
F = \frac{G \cdot M \cdot m}{r^2}
\]

Where:
- \( F \) is the gravitational force (the man's weight),
- \( G \) is the gravitational constant, \( 6.67408 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2} \),
- \( M \) is the mass of the Earth, \( 5.9742 \times 10^{24} \, \text{kg} \),
- \( m \) is the man's mass (which we can find from his weight on the surface of the Earth),
- \( r \) is the distance from the center of the Earth to the man.

### Step 1: Find the man's mass
We are given that the man weighs 150 N on the Earth's surface. Using the gravitational force formula on the Earth's surface:

\[
150 = \frac{G \cdot M \cdot m}{R_E^2}
\]

Where \( R_E \) is the radius of the Earth, \( 6378 \, \text{km} = 6.378 \times 10^6 \, \text{m} \). Solving for the man's mass \( m \):

\[
m = \frac{150 \cdot R_E^2}{G \cdot M}
\]

Substitute the values:

\[
m = \frac{150 \cdot (6.378 \times 10^6)^2}{6.67408 \times 10^{-11} \cdot 5.9742 \times 10^{24}}
\]

Let me calculate that.

The mass of the man is approximately \( 15.30 \, \text{kg} \).

### Step 2: Calculate the weight at an elevation equal to the radius of the Earth

At an elevation equal to the Earth's radius, the total distance from the center of the Earth becomes twice the Earth's radius:

\[
r = 2 \cdot R_E = 2 \cdot 6.378 \times 10^6 \, \text{m} = 1.2756 \times 10^7 \, \text{m}
\]

Now, we can calculate his weight at this new distance using the gravitational force formula:

\[
F' = \frac{G \cdot M \cdot m}{r^2}
\]

Substituting the values:

\[
F' = \frac{6.67408 \times 10^{-11} \cdot 5.9742 \times 10^{24} \cdot 15.30}{(1.2756 \times 10^7)^2}
\]

Let me compute this value.At an elevation equal to the radius of the Earth, the man's weight would be approximately **37.5 N**.

This is significantly less than his weight on the Earth's surface (150 N), due to the increased distance from the Earth's center.

Would you like further details, or do you have any other questions? 

Here are 5 related questions you might find interesting:
1. How does the gravitational force change with increasing distance?
2. What would happen to the man's weight at an elevation equal to twice the Earth's radius?
3. How does the mass of an object affect its gravitational pull on other objects?
4. What would the man's weight be on the surface of the Moon, given the Moon's mass and radius?
5. How does the gravitational force between two objects depend on their masses and the distance between them?

**Tip:** Gravitational force decreases with the square of the distance. So, doubling the distance reduces the force to one-quarter of its original value.

A man weighs 150 N on the surface of the Earth. Find his weight at an elevation equal to the radius of the Earth. Mass of Earth = 5.9742 x 10^24 kg, Radius of Earth = 6378 km, Mass of Moon = 0.073483 x 10^24 kg, Mass of the Sun = 1.9891 × 10^30 kg, Distance between Earth and Sun = 149.6 x 10^6 km, G = 6.67408 x 10^-11 m^3kg^-1s^-2.

Find the weight of a man at an elevation equal to the Earth's radius using gravitational force calculations. This example demonstrates key physics concepts such as Newton's Law of Universal Gravitation, inverse square law, and gravitational constants. Perfect for Grades 10-12 physics students.

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