Chapter 6: Problem 745
3 particle each of mass \(\mathrm{m}\) are kept at vertices of an equilateral triangle of side \(L\). The gravitational field at center due to these particles is (A) zero (B) \(\left[(3 \mathrm{GM}) / \mathrm{L}^{2}\right]\) (C) \(\left[(9 \mathrm{GM}) / \mathrm{L}^{2}\right]\) (D) \((12 / \sqrt{3})\left(\mathrm{Gm} / \mathrm{L}^{2}\right)\)
Short Answer
Step by step solution
Gravitational field formula
Calculate the gravitational field from each particle
Sum the gravitational fields
Finalize the answer and choose the correct option
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Newton's Law of Gravitation
- \[ F = G \frac{m_1 m_2}{r^2} \]
Equilateral Triangle
- Each particle will pull on the centroid equally in terms of direction and magnitude.
- The symmetry results in simplified calculations, such as noting that the forces or fields cancel out.
Vector Addition
- This angle is fundamental, as the cosine of 120 degrees is negative, affecting how vectors combine.
- Using the cosine rule, we calculate the magnitude of the total gravitational field by considering both magnitudes and angles.
Gravitational Constant
- It allows us to compute the gravitational field using the formula \( g = G \frac{m}{r^2} \).
- The constant is universal, making its application possible across all gravitational calculations regardless of the masses or distances involved.