A missile weighing 400N on the earth surface is shot into the atmosphere to an altitude of 6.4 x \(10^{-6}\)m. Taking the earth as a sphere of radius 6.4 x \(10^{-6}\)m and assuming the inverse-square law of universal gravitation, what would be the weight of the missile at that altitude?
To find the weight of the missile at an altitude using the inverse-square law of gravitation, we can use the formula for gravitational force:
\(F = \frac{G \cdot m_1 \cdot m_2}{r^2}\)
where:
- \( F \) is the gravitational force (weight),
- \( G \) is the gravitational constant,
- \( m_1 \) is the mass of the Earth,
- \( m_2 \) is the mass of the missile,
- \( r \) is the distance from the center of the Earth to the missile.
Step 1: Determine the radius at altitude
1. The radius of the Earth is \( R = 6.4 \times 10^6 \, \text{m} \).
2. The altitude \( h = 6.4 \times 10^6 \, \text{m} \).
Thus, the distance from the center of the Earth to the missile at that altitude is:
\(r = R + h = 6.4 \times 10^6 \, \text{m} + 6.4 \times 10^6 \, \text{m} = 12.8 \times 10^6 \, \text{m}\)
Step 2: Calculate the weight at the altitude
The weight of the missile at the surface is given as \( W_0 = 400 \, \text{N} \). The weight at altitude can be calculated using the ratio of the squares of the distances:
\(W = W_0 \cdot \left( \frac{R}{r} \right)^2\)
Substituting the values:
\(W = 400 \cdot \left( \frac{6.4 \times 10^6}{12.8 \times 10^6} \right)^2\)
Calculating this:
\(W = 400 \cdot \left( \frac{1}{2} \right)^2 = 400 \cdot \frac{1}{4} = 100 \, \text{N}\)
The weight of the missile at an altitude of \( 6.4 \times 10^6 \, \text{m} \) is \( 100 \, \text{N} \).
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