If the weight of a body on the Earth's surface is 300 N, calculate the height from the Earth's surface at which the body's weight measures 200 N. (The radius of the Earth is 6380 km) [1.43×10^0 m]



Answer :

Sure, let's solve this step by step using the principles of gravitational force.

We are given:
- The weight of the body on Earth's surface (W₁) = 300 N
- The weight of the body at some height above the Earth's surface (W₂) = 200 N
- The radius of the Earth (R₁) = 6380 km

First, let's convert the radius of the Earth from kilometers to meters:
[tex]\[ R₁ = 6380 \, \text{km} \times 1000 \, \text{m/km} = 6,380,000 \, \text{m} \][/tex]

We know that gravitational force (and thereby weight) depends on the inverse square of the distance from the center of the Earth. The formula for weight at a distance (r) from the center of the Earth is:

[tex]\[ W = \frac{G M m}{r^2} \][/tex]

where:
- [tex]\( G \)[/tex] is the gravitational constant,
- [tex]\( M \)[/tex] is the mass of the Earth,
- [tex]\( m \)[/tex] is the mass of the object,
- [tex]\( r \)[/tex] is the distance from the center of the Earth.

We can set the ratio of the weights given for the two distances:
[tex]\[ \frac{W_2}{W_1} = \frac{R_1^2}{(R_1 + h)^2} \][/tex]

where [tex]\( h \)[/tex] is the height above the Earth's surface where the weight measures 200 N.

Given:
[tex]\[ \frac{200 \, \text{N}}{300 \, \text{N}} = \frac{R_1^2}{(R_1 + h)^2} \][/tex]

Simplify this ratio:
[tex]\[ \frac{2}{3} = \frac{R_1^2}{(R_1 + h)^2} \][/tex]

Cross-multiplying to solve for [tex]\( R_1 + h \)[/tex]:
[tex]\[ \frac{R_1^2}{(R_1 + h)^2} = \frac{2}{3} \][/tex]
[tex]\[ R_1^2 = \frac{2}{3}(R_1 + h)^2 \][/tex]

Taking the square root of both sides:
[tex]\[ \sqrt{\frac{2}{3}}R_1 = R_1 + h \][/tex]

Solving for [tex]\( h \)[/tex]:
[tex]\[ R_1 \sqrt{\frac{2}{3}} = R_1 + h \][/tex]
[tex]\[ h = R_1 \left(\sqrt{\frac{2}{3}} - 1\right) \][/tex]

Plugging in the value of radius [tex]\( R_1 \)[/tex]:

[tex]\[ h = 6,380,000 \, \text{m} \left(\sqrt{\frac{2}{3}} - 1\right) \][/tex]

And calculating this:

[tex]\[ h \approx 1,433,872.2794783385 \, \text{m} \][/tex]

Therefore, the height from the Earth's surface at which the weight of the body would be 200 N is approximately:

[tex]\[ 1.43 \times 10^6 \, \text{m} \][/tex]