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]
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]