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in this chart, the column headers identify a change to mass 2, and the …

Question

in this chart, the column headers identify a change to mass 2, and the row headers identify a simultaneous change to the distance between mass 1 and mass 2. for all cases, mass 1 is unchanged
determine the force between m1 and m2 due to the changes in both distance and mass 2. you will drag and drop the appropriate fraction to indicate how the force has been changed. the row where r is unchanged has been completed for you.
i need a hint

Explanation:

Step1: Recall Newton's law of universal gravitation

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

Step2: Analyze the case when \(r=\frac{r}{2}\) and \(m_2 = \frac{m}{2}\)

Substitute into the formula: \(F'=G\frac{m_1\times\frac{m}{2}}{(\frac{r}{2})^{2}}=G\frac{m_1m}{2}\times\frac{4}{r^{2}} = 2G\frac{m_1m}{r^{2}} = 2F\)

Step3: Analyze the case when \(r = 2r\) and \(m_2=\frac{m}{2}\)

Substitute into the formula: \(F'=G\frac{m_1\times\frac{m}{2}}{(2r)^{2}}=G\frac{m_1m}{4r^{2}}\times\frac{1}{2}=\frac{1}{8}G\frac{m_1m}{r^{2}}=\frac{1}{8}F\)

Step4: Analyze the case when \(r = 3r\) and \(m_2=\frac{m}{2}\)

Substitute into the formula: \(F'=G\frac{m_1\times\frac{m}{2}}{(3r)^{2}}=G\frac{m_1m}{9r^{2}}\times\frac{1}{2}=\frac{1}{18}G\frac{m_1m}{r^{2}}=\frac{1}{18}F\)

Step5: Analyze the case when \(r = 2r\) and \(m_2 = m\)

Substitute into the formula: \(F'=G\frac{m_1\times m}{(2r)^{2}}=\frac{1}{4}G\frac{m_1m}{r^{2}}=\frac{1}{4}F\)

Step6: Analyze the case when \(r = 3r\) and \(m_2 = m\)

Substitute into the formula: \(F'=G\frac{m_1\times m}{(3r)^{2}}=\frac{1}{9}G\frac{m_1m}{r^{2}}=\frac{1}{9}F\)

Step7: Analyze the case when \(r = 2r\) and \(m_2 = 2m\)

Substitute into the formula: \(F'=G\frac{m_1\times2m}{(2r)^{2}}=\frac{1}{2}G\frac{m_1m}{r^{2}}=\frac{1}{2}F\)

Step8: Analyze the case when \(r = 3r\) and \(m_2 = 2m\)

Substitute into the formula: \(F'=G\frac{m_1\times2m}{(3r)^{2}}=\frac{2}{9}G\frac{m_1m}{r^{2}}=\frac{2}{9}F\)

Answer:

When \(r=\frac{r}{2}\):

  • For \(m_2=\frac{m}{2}\), \(A = 2F\)

When \(r = 2r\):

  • For \(m_2=\frac{m}{2}\), \(D=\frac{1}{8}F\); for \(m_2 = m\), \(E=\frac{1}{4}F\); for \(m_2 = 2m\), \(F=\frac{1}{2}F\)

When \(r = 3r\):

  • For \(m_2=\frac{m}{2}\), \(G=\frac{1}{18}F\); for \(m_2 = m\), \(H=\frac{1}{9}F\); for \(m_2 = 2m\), \(I=\frac{2}{9}F\)