Question

Please help

Suppose the force between the Earth and Moon were electrical instead of gravitational, with the Earth having a positive charge and the Moon having a negative one. If the magnitude of each charge were proportional to the respective body’s mass, find the Q/m ratio required for the moon to follow its present orbit of 3.84 x 108 m radius with its period of 27.3 days. The Earth’s mass is 5.98 x 1024 kg, and the moon’s mass is 7.3 x 1022 kg.

Answer 1

Calculate the centripetal force required first.

Answer 2

would I make it equal to the electric force after?

Answer 3

yes

Answer 4

would I have to do anything after that?

Answer 5

also would the mass of earth and moon be the charges on them?

Answer 6

No. You'll have
\[Q_{moon} = C \cdot M_{moon} \]
and \[Q_{earth} = C \cdot M_{earth} \]
so
\[Q_{moon} Q_{earth} = C^2 M_{moon} M_{earth} \]
the question wants you to solve for C.

Answer 7

so the charge of the moon is equal to the centripetal force multiplied by the mass of the moon

Answer 8

No, of course not. The C is just a constant.

Answer 9

so I will isolate for c then

Answer 10

yes

Answer 11

so I will isolate c first, the find centripetal force then electric

Answer 12

I'm not sure what you mean, but just try it.

Answer 13

so when I find C, the constant, will that be the answer? because I don't know were finding centripetal force and electrical force come in

Answer 14

Yes it will. You can find the centripetal force and set that equal to the electric force. That's one equation, and the unknown is C.

Answer 15

so the equation will be

Me Mm 4pirf^2 = (kQ1Q2/ r^2) C

Answer 16

No.

Answer 17

What is the expression for centripetal force?

Answer 18

v^2/r = 4pi^2 rf^2=4pi^2r/T^2

Answer 19

That's centripetal acceleration. You need to multiply by the mass of the object that's moving in the circle.

Answer 20

I forgot the mass in each equation tho.

Answer 21

so it will be Mm 4pirf^2 = (kQ1Q2/ r^2) C , but Mm will be only the mass of the moon

Answer 22

The left side is right. The right side has no C in it. The C comes in when you replace the Q's with M's, like I did above.

Answer 23

so with that equation I will solve for Q for the earth and moon

Answer 24

Just read what I wrote above. Qe Qm = C^2 Me Mm

Answer 25

ok, I will try it then then come back here to see if I'm doing it right and if I get what you mean

Answer 26

so i found Q for the moon, but for the earth i don't have r or f so i don't know how to do it. i used the equation Q= m4pi^2r^3f^2/k

Answer 27

\[F_c = 4\pi^2M_{moon}fR = k \frac{Q_{earth}Q_{moon} }{R^2}= C^2k \frac{M_{earth}M_{moon}}{R^2}\]
Can you do it from here?

Answer 28

yes. sorry

Answer 29

Quite alright.

Answer 30

ugh I'm very sorry. But what is messing me up is the Q and M. Because I don't know what I have to sub in for Q, and solve for. I know I have to solve for C for the last equation, but what about the other two, am I solving for Fe for the other two and making it equal to the one with c and isolating c

Answer 31

You are setting the first equal to the last. You know all of those values except for C.

Answer 32

Well, the second -- the one that starts with 4 pi ^2