An asteroid of mass 58,000 kg carrying a negative charge of 15E-6 C...

Question

An asteroid of mass 58,000 kg carrying a negative charge of 15E-6 C is 180 m from a second asteroid of mass 52,000 kg carrying a negative charge of 11E-6 C. The magnitude of the net force the asteroids exert upon each other, assuming we can treat them as point particles is  Blank 1. Fill in the blank, read surrounding text. N.  

Answer 1

This problem combines two different concepts: gravity and electrostatics. Because these asteroids have mass, they are gravitationally attracted to each other. Because they both have negatively charged, they are electrically repelled from each other. Our strategy is to find both of these forces and add them together.

Gravitational Force

Newton's Universal Law of Gravitation is,

F_G = GM₁M₂/r²

where M₁ and M₂ are the masses of asteroids 1 and 2, r is the distance of separation between the two points, and G is the gravitational constant, a fundamental constant whose value is 6.67x10^-11 Nm²/kg².

In our case, we have M₁ = 58,000 kg, M₂ = 52,000 kg, and r = 180 m. Plugging this in, we get:

F_G = GM₁M₂/r² = (6.67x10^-11 Nm²/kg²)(58000 kg)(52000 kg)/((180 m)^2)

= 6.2089x10^-6 N

= 6.2089 µN of gravitational force

Electrostatics

Coulomb's Law is,

F_E = k_eq₁q₂/r²

where q₁ and q₂ are charges of asteroids 1 and 2, r is the distance of separation between the two points, and k_e is Coulomb's constant, where k_e = 1/(4πε₀) = 8.99x10⁹ Nm²/C².

In our case we have q₁ = 15x10^-6 C, q₂ = 11x10^-6 C, and r = 180 m. Plugging this in, we get:

F_E = k_eq₁q₂/r² = (8.99x10⁹ Nm²/C²)(15x10^-6 C)(11x10^-6 C)/((180 m)^2)

= 45.7824x10^-6 N

= 45.7824 µN

Putting Things Together

Finally, we add these forces together to find the final resultant force. Remember, gravitational forces are always attractive, but our like-charge electrostatic force here (negative-negative) is repulsive. We sum the two forces, but since they point in different directions, we have to give them different signs.

Let's choose the attractive force to be positive and the repulsive force to be negative. Then we have our sum of forces to be:

F_Total = ΣF = F_G + F_E

F_Total = 6.2089 µN + (-45.7824 µN)

F_Total = -39.5735 µN

|F_Total| = 39.5735 µN

Our sign is negative, meaning our force is repulsive, and we've found it's magnitude to be 39.5735 µN.