Chapter 9: Problem 20
A particle hanging from a massless spring stretches it by \(2 \mathrm{~cm}\) at earth's surface. How much will the same particle stretch the spring at height \(2624 \mathrm{~km}\) from the surface of earth ? (Radius of earth \(=6400 \mathrm{~km}\) ) (a) \(1 \mathrm{~cm}\) (b) \(2 \mathrm{~cm}\) (c) \(3 \mathrm{~cm}\) (d) \(4 \mathrm{~cm}\)
Short Answer
Step by step solution
Understand the Problem
Model Gravitational Force at Earth's Surface
Calculate Gravitational Force at Height
Calculate Stretch at Height
Simplify and Solve
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gravitational Force
- \( F = \frac{G \, m_1 \, m_2}{r^2} \)
- \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{N} \, (\text{m/kg})^2) \),
- \( m_1 \) and \( m_2 \) are the masses of the objects, and
- \( r \) is the distance between the centers of the two masses.
Acceleration Due to Gravity
- \( g' = g \left( \frac{R}{R + h} \right)^2 \)
Spring Extension
- \( F = kx \)
- \( x = \frac{F}{k} \)
Earth's Radius
- \( g' = g \left( \frac{R}{R + h} \right)^2 \),