Question

N asteroid of mass 58,000 kg carrying a negative charge of 15 μc is 180 m from a second asteroid of mass 52,000 kg carrying a negative charge of 11 μc. what is the magnitude of the net force the asteroids exert upon each other, assuming we can treat them as point particles? (g = 6.67 × 10-11 n • m2/kg2, k = 1/4 πε0 = 8.99 × 109 n • m2/c2)

Answer 1

For this problem, we use the Coulomb's Law. The working equation is written below:

F = kQ₁Q₂/d²
where 
F is the electric force
k is a constant equal to 8.99 × 10⁹ N • m²/C²
Q is the charges for the two objects
d is he distance between the objects

Substituting the values,
F = (8.99 × 10⁹ N • m²/C²)(-15×10⁻⁶ C)(-11×10⁻⁶ C)/(180²)
F = 4.578×10⁻⁵ N

Next, we determine the gravitational force using the Law of Universal Gravitation:

F = Gm₁m₂/d²
where
F is the gravitational force
G is a constant equal to 6.67 × 10⁻¹¹ N • m²/kg²
m is the masses of the objects
d is the distance between the objects

F = (6.67 × 10⁻¹¹ N • m²/kg²)(58,000 kg)(52,000 kg)/(180²)
F = 6.2089×10⁻⁶ N

The sum of the two forces equal the net force:
Net force = 4.578×10⁻⁵ N + 6.2089×10⁻⁶ N = 1.079×10⁻⁵ N 

Answer 2

Answer: The net force between the the asteroids will $-3.9492\times 10^{-5} N$ and the negative sign indicates that they repel each other.

Explanation:

The Coulombs forces exerted by asteroids:

Charge on asteroid-1 = $Q_1=-15 \mu C=-15\times 10^{-6} C$

Charge on asteroid-2 = $Q_1=-11 \mu C=-11\times 10^{-6} C$

Distance between the asteroids = 180 m

$k=\frac{1}{4\pi\epsilon _o}=8.99\times 10^9 N m^2/c^2$

$F_c=k\times \frac{Q_1\times Q_2}{r^2}=8.99\times 10^9 N m^2/c^2\times \frac{-15\times 10^{-6} C\times -11\times 10^{-6} C}{(180 m)^2}=4.57\times 10^{-5} N$

Since, the charges are negative they will repel each other.

The gravitational force between the asteroids:

Mass of the asteroid -1 = $m_1$ = 58000 kg

Mass of the asteroid -2 = $m_1$ = 52000 kg

G = $6.67\times 10^{-11}N m^2/kg^2$

$F_G=G\times \frac{m_1\times m_2}{r^2}=6.67\times 10^{-11}N m^2/kg^2\times \frac{58,000 kg\times 52000 kg}{(180 m)^2}=6.208\times 10^{-6} N$

The gravitational force is an attractive force.

Since, the charges are negatively charged they will repel each other.

The net force = $F=F_G+(-F_C)=6.208\times 10^{-6} N+(-4.57\times 10^{-5} N)=-3.9492\times 10^{-5} N$

Hence, the net force between the the asteroids will $-3.9492\times 10^{-5} N$ and the negative sign indicates that they repel each other.