Question

force of gravity vs. mass
fill in the table below for the values of
gravitational force.

Explanation:

Step1: Recall the gravitational force formula

The gravitational force formula is \(F = G\frac{m_1m_2}{r^2}\). Assuming the distance \(r\) and the gravitational constant \(G\) are constant, we can consider the ratio of forces based on the product of masses \(m_1m_2\). Let the initial force be \(F_0=G\frac{m_1m_2}{r^2}\).

Step2: Calculate for \(m_1 = 2m_1\) and \(m_2=\frac{1}{4}m_2\)

The new product of masses is \(m_{1\text{new}}m_{2\text{new}}=(2m_1)\times(\frac{1}{4}m_2)=\frac{1}{2}m_1m_2\). Since \(F = k m_1m_2\) (where \(k = \frac{G}{r^2}\) is constant), the new force \(F_1=\frac{1}{2}F_0\).

Step3: Calculate for \(m_1=\frac{1}{2}m_1\) and \(m_2 = 2m_2\)

The new product of masses is \(m_{1\text{new}}m_{2\text{new}}=(\frac{1}{2}m_1)\times(2m_2)=m_1m_2\). So the new force \(F_2 = F_0\).

Step4: Calculate for \(m_1=\frac{1}{4}m_1\) and \(m_2 = 10m_2\)

The new product of masses is \(m_{1\text{new}}m_{2\text{new}}=(\frac{1}{4}m_1)\times(10m_2)=\frac{10}{4}m_1m_2=\frac{5}{2}m_1m_2\). Thus the new force \(F_3=\frac{5}{2}F_0\).

Step5: Calculate for \(m_1 = 10m_1\) and \(m_2=\frac{1}{2}m_2\)

The new product of masses is \(m_{1\text{new}}m_{2\text{new}}=(10m_1)\times(\frac{1}{2}m_2)=5m_1m_2\). So the new force \(F_4 = 5F_0\).

Answer:

\(m_1\)	\(m_2\)	\(F\)
\(\frac{1}{2}m_1\)	\(2m_2\)	\(F_0\)
\(\frac{1}{4}m_1\)	\(10m_2\)	\(\frac{5}{2}F_0\)
\(10m_1\)	\(\frac{1}{2}m_2\)	\(5F_0\)