Question

set the mass of both objects to $10.0 \cdot 10^5$ kg and check the \show distance\ box at the top of the simulation.
drag object a to $(-5, 0)$ and object b to $(5, 0)$.
what is the distance between the two objects and the force felt by each? enter your answers in the boxes.
$d = \square$ m
$|f_a| = |f_b| = \square$ n

Explanation:

Step1: Calculate the distance between two points

The coordinates of object A are \((-5, 0)\) and object B are \((5, 0)\). The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here, \(y_1 = y_2 = 0\), so \(d=\vert5 - (-5)\vert=\vert10\vert = 10\) m.

Step2: Calculate the gravitational force between the two objects

The gravitational force formula is \(F = G\frac{m_1m_2}{r^2}\), where \(G = 6.674\times10^{-11}\space Nm^2/kg^2\), \(m_1=m_2 = 10.0\times10^{5}\space kg\), and \(r = d = 10\space m\).
Substitute the values into the formula:
\(F=6.674\times10^{-11}\times\frac{(10.0\times10^{5})\times(10.0\times10^{5})}{10^{2}}\)
First, calculate the numerator: \((10.0\times10^{5})\times(10.0\times10^{5})=100\times10^{10}=1\times10^{12}\)
Then, the denominator is \(10^{2} = 100\)
So \(F = 6.674\times10^{-11}\times\frac{1\times10^{12}}{100}=6.674\times10^{-11}\times1\times10^{10}= 0.6674\space N\) (approximately \(0.667\) or \(0.67\) depending on significant figures, but using the exact calculation with \(G = 6.67\times10^{-11}\) we get \(F=6.67\times10^{-11}\times\frac{(10^{6})\times(10^{6})}{100}=6.67\times10^{-11}\times\frac{10^{12}}{100}=6.67\times10^{-11}\times10^{10}=0.667\space N\))

Answer:

For the distance \(d\): \(10\) m.
For the force \(\vert F_A\vert=\vert F_B\vert\): approximately \(6.67\times 10^{-1}\space N\) (or \(0.667\space N\))