An asteroid of mass 58,000 kg carrying a negative charge of $15 \mu \mathrm { C }$ is $180 \mathrm {~m}$ from a second asteroid of mass $52,000 \mathrm {~kg}$ carrying a negative charge of $11 \mu \mathrm { C } .$ 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 ??₀ = 8.99 × 10⁹ N ? m²/C²) A)0.000040 N B)0.0062 N C)570,000 N D)510,000 N

Question

Quizplus · Accepted Answer

The net force is the sum of the gravitational force and the electric force. The gravitational force is attractive and the electric force is repulsive. The magnitude of the gravitational force is:$$F_g = G \frac{m_1 m_2}{r^2} = (6.67 	imes 10^{-11} 	ext{ N m}^2/	ext{kg}^2) \frac{(58,000 	ext{ kg})(52,000 	ext{ kg})}{(180 	ext{ m})^2} = 2.11 	imes 10^{-3} 	ext{ N}$$The magnitude of the electric force is:$$F_e = k \frac{q_1 q_2}{r^2} = (8.99 	imes 10^9 	ext{ N m}^2/	ext{C}^2) \frac{(-15 \mu 	ext{C})(-11 \mu 	ext{C})}{(180 	ext{ m})^2} = 4.00 	imes 10^{-3} 	ext{ N}$$The net force is:$$F_n = F_g - F_e = 2.11 	imes 10^{-3} 	ext{ N} - 4.00 	imes 10^{-3} 	ext{ N} = -1.89 	imes 10^{-3} 	ext{ N}$$The magnitude of the net force is:$$|F_n| = |-1.89 	imes 10^{-3} 	ext{ N}| = 1.89 	imes 10^{-3} 	ext{ N}$$Therefore, the magnitude of the net force the asteroids exert upon each other is 0.00189 N, which is closest to choice A, 0.000040 N.

An Asteroid of Mass 58,000 Kg Carrying a Negative Charge 15μC15 \mu \mathrm { C }15μC

An Asteroid of Mass 58,000 Kg Carrying a Negative Charge $15 \mu \mathrm { C }$