## anonymous 4 years ago Any idea where to start for this problem, I really have no idea : Supose that earth and the moon have positive charges of the same magnitude, how big does the charge need to be to generate a electrostatic repulsion equal to 1.00% of the gravitational attraction between the bodies? thank you so much!

1. TuringTest

(I'm assuming you mean 100%, though you wrote 1.00%=1%) We have two forces here that are going to be equal in magnitude: gravity and charge. The formula for gravity between two masses m1 and m2 is$F_g=G\frac{m_1m_2}{r^2}$The formula for electrostatic force between two charges q1 and q2 is$F_e=k\frac{q_1q_2}{r^2}$we are told here that the charge on the two bodies in question (the earth and moon) are equal, so q1=q2 in this case. Setting the forces equal we get$F_g=F_e$$G\frac{m_{earth}m_{moon}}{r^2}=k\frac{q^2}{r^2}$which we can solve for q easily. If your question was in fact supposed to be that the charges create 1% of the gravitational force, we would simply alter our starting equation to$\frac{F_g}{100}=F_e$and do the same as above.

2. TuringTest

...the value of the constants k and G are well-known and can be found online, by the way.

3. anonymous

It is 1%, I dont know why the teacher used this notation 1.00% , so then it would just be the result of $(k(q^2/r^2)) 100 \right?$

4. TuringTest

can't read what you wrote in latex very well, but F_e=1% of F_g leads to$100k\frac{q^2}{r^2}=G\frac{m_1m_2}{r^2}$in our case. I think that is what you wrote above...

5. anonymous

hah yep, thank you so much! for r should I use the distance between them?

6. TuringTest

Normally yes, but look at the two formulas when set equal and notice that the r^2's will cancel. So it really doesn't matter here.