Question

An asteroid of mass 5.7×104 kg carrying a negative charge of 14.0 μC is 190 m from a second asteroid of mass 5.5×104 kg carrying a negative charge of 15.0 μC. What is the net force the asteroids exert upon each other? (G = 6.673 x 10-11 N∙m2/kg2 and k = 9.0 x 109 N∙m2/C2.) An asteroid of mass 5.7×104 kg carrying a negative charge of 14.0 μC is 190 m from a second asteroid of mass 5.5×104 kg carrying a negative charge of 15.0 μC. What is the net force the asteroids exert upon each other? (G = 6.673 x 10-11 N∙m2/kg2 and k = 9.0 x 109 N∙m2/C2.) 5.6×105 N 5.4×105 N 5.7×10−3 N 4.7×10−5 N

Answer 1

repulsion force es R = 5,224 10⁻³ N

Step-by-step explanation:

We use Newton's second law to add forces, the elective force is repulsive because the asteroids have the same type of charge and gravitational force is always attractive.

Gravitational force

Fg = G m1 m2 / r²

Fg = 6,673 10⁻¹¹ 5.7 10⁴ 5.5 10⁴/190²

Fg = 5,795 10⁻⁶ N

Electric force

Fe = k q1 q2 / r²

Fe = 8.99 10⁹ 14.0 10⁻⁶ 15.0 10⁻⁶ / 190²

Fe = 5.2296 10⁻³ N

The resulting force is

R = Fe - Fg

R = 5.2296 10⁻³ - 5.795 10⁻⁶= 5229.6 10⁻⁶ - 5,795 10⁻⁶

R = 5,224 10⁻³ N

this is a repulsive force and the asteroids move away from each other