WAEC - Physics (1997 - No. 6)

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?
100N
200N
400N
800N
1600N

Explanation

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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