What Charge Value Would Keep the Moon in Orbit by Electrical Force?

[acissej2487]

Here is my problem:

Suppose that electrical attraction, rather than gravity, were responsible for holding the Moon in orbit around the Earth.
If equal and opposite charges were placed on the Earth and the Moon, what should be the value of Q to maintain the present orbit? Use these data: mass of Earth=5.98*10^24 kg, mass of Moon = 7.35*10^22 kg, radius of orbit=3.84*10^8 m. Treat the Earth and Moon as point particles.


I know that I use Gm1m2/r^2 = kQ2/r^2. But I cannot get the right answer!

Help please :)

 

[rl.bhat]

I know that I use Gm1m2/r^2 = kQ2/r^2. But I cannot get the right answer!
You have to use m2v^2/r = kQ2/r^2 where m2 is the mass of the moon and v is the velocity of the moon around the earth.

 

[thomate1]

Thank you for presenting your problem. It is an interesting concept to consider electrical attraction as the force responsible for holding the Moon in orbit around the Earth. To solve this problem, we can use the equation for Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this case, we can consider the Earth and Moon as two point particles with equal and opposite charges, and the force between them is equal to the gravitational force that keeps the Moon in orbit.

Using the given data, we can calculate the gravitational force between the Earth and Moon as:

F = G(m1m2)/r^2

Where G is the gravitational constant, m1 and m2 are the masses of the Earth and Moon respectively, and r is the radius of the orbit.

Plugging in the values, we get:

F = (6.67*10^-11 Nm^2/kg^2) * (5.98*10^24 kg * 7.35*10^22 kg) / (3.84*10^8 m)^2 = 1.99*10^20 N

Now, since we know that the electrical force between the two particles must be equal to this gravitational force, we can set up the equation:

F = k(Q^2)/r^2

Where k is the Coulomb's constant and Q is the charge on each particle.

Plugging in the known values for k, F, and r, we can solve for Q:

1.99*10^20 N = (8.99*10^9 Nm^2/C^2)(Q^2)/(3.84*10^8 m)^2

Solving for Q, we get:

Q = 3.34*10^-5 C

Therefore, if the Earth and Moon had equal and opposite charges of 3.34*10^-5 C, the electrical force between them would be equal to the gravitational force that keeps the Moon in orbit.

I hope this helps to solve your problem. Keep up the good work in exploring different scientific concepts!