Question

The gravitational force of attraction be-
tween two students sitting at their desks in
physics class is 3.47 x 10-8 N.
If one student has a mass of 44.8 kg and the other has a mass of 53.9 kg, how far apart are the students sitting? The universal gravitational constant is 6.673 x 10 11 N m² kg?.
Answer in units of m.

please help!! :)

Answer 1

Answer:

2.15 m

Explanation:

Newton's Law of Universal Gravitation:

$\displaystyle F_g = G \frac{m_1 m_2}{r^2}$  

• $\displaystyle F_g$ is the gravitational force of attraction

• $\displaystyle G$ is the universal gravitational constant

• $m_1$ and $m_2$ are the two masses of the two objects

• $\displaystyle r$ is the distance between the centers of the two objects.

List the known values:

• $\displaystyle F_g = 3.47 \cdot 10^-^8 \ \text{N}$  
• $\displaystyle G = 6.673 \cdot 10^-^1^1 \ \frac{Nm^2}{kg^2}$
• $\displaystyle m_1 = 44.8 \ \text{kg} \\ m_2 = 53.9 \ \text{kg}$
• $\displaystyle r =\ ?$

Plug these values into the equation:

• $\displaystyle 3.47 \cdot 10^-^8 \ \text{N} = 6.673 \cdot 10^-^1^1 \ \frac{Nm^2}{kg^2} \frac{(44.8 \ \text{kg})(53.9 \ \text{kg})}{r^2}$

Notice that the units $\displaystyle \text{N}$, $\displaystyle \text{kg}^2$, and $\displaystyle \text{m}$ cancel out. We are left with the unit $\displaystyle \text{m}$ for radius r.

Get rid of the units to make the problem easier to read.

• $\displaystyle 3.47 \cdot 10^-^8= 6.673 \cdot 10^-^1^1 \ \frac{(44.8)(53.9) \ }{r^2}$  

Multiply the masses together.

• $\displaystyle 3.47 \cdot 10^-^8= 6.673 \cdot 10^-^1^1 \ \frac{(2414.72)}{r^2}$

Multiply the gravitational constant and the masses together.

• $\displaystyle 3.47 \cdot 10^-^8= \frac{1.61134265\cdot 10^-^7}{r^2}$

Solve for $r^2$ by dividing both sides by 3.47 * 10^(-8) and moving $r^2$ to the left.

• $\displaystyle r^2 = \frac{1.61134265\cdot 10^-^7}{3.47\cdot 10^-^8}$
• $r^2 = 4.64363876$

Take the square root of both sides.

• $r=2.154910383$  

The students are sitting about 2.15 m apart from each other.