Question

if m₁ is doubled, and m₂ is reduced by 3x, what will be the gravitational force between the two object?

Explanation:

Step1: Write the formula for gravitational force

The formula for gravitational force is \(F = G\frac{m_1m_2}{r^2}\), where \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the two objects, and \(r\) is the distance between them.

Step2: Substitute the new values of \(m_1\) and \(m_2\)

If \(m_1\) is doubled (\(m_1'= 2m_1\)) and \(m_2\) is reduced by \(3x\) (assuming it means divided by \(3\), \(m_2'=\frac{m_2}{3}\)), the new force \(F'\) is \(F'=G\frac{(2m_1)(\frac{m_2}{3})}{r^2}\).

Step3: Simplify the expression

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Since \(F = G\frac{m_1m_2}{r^2}\), then \(F'=\frac{2}{3}F\)

Answer:

The new gravitational force will be \(\frac{2}{3}\) of the original gravitational force.