Question

Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible compared to the gravitational attraction between the ships. Construct a problem in which you place identical excess charges on the space ships to exactly counter their gravitational attraction. Calculate the amount of excess charge needed. Examine whether that charge depends on the distance between the centers of the ships, the masses of the ships, or any other factors. Discuss whether this would be an easy, difficult, or even impossible thing to do in practice.

Answer 1

q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Step-by-step explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

F_e -F_g = 0

F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

`k (q_1q_2)/(r^2) = G (m_1m_2)/(r^2)`

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

k q² = G m²

q =
`\sqrt{ (G)/(k) }` m

we substitute

q =
`\sqrt{ ( 6.67 \ 10^(-11))/(8.99 \ 10^(9)) }` m

q =
`\sqrt{0.7419 \ 10^(-20)}` m

q = 0.861 10⁻¹⁰ m

q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

q = 8.61 10⁻¹¹ 10³

q = 8.61 10⁻⁸ C

this is a very small charge value so it should be easy to create in each one