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41P

Expert-verifiedFound in: Page 627

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: (a) what equal positive charges would have to be placed on Earth and on the Moon to neutralize their gravitational attraction? (b) Why don’t you need to know the lunar distance to solve this problem? (c) How many kilograms of hydrogen ions (that is, protons) would be needed to provide the positive charge calculated in (a)?**

- The value of equal positive charges to be placed on Earth and on the Moon to neutralize their gravitational attraction is$5.7\times {10}^{13}C$.
- As the distance gets cancelled due to both their electric and gravitational charges, we don’t need to know the lunar distance to solve this problem.
- The amount of hydrogen ions needed to provide the positive charge calculated in (a) is $6.0\times {10}^{5}kg$

Positive charges are needed to be placed on Earth and on the moon to neutralize their gravitational attraction.

**Using the concept of both gravitational and electrostatic force from Coulomb's law, we can get the value of the positive charge of the particle. Again, we can get the number of hydrogen ions using the concept of the net charge**

Formulae:

The magnitude of the electrostatic force between any two particles$F=k\frac{\left|{q}_{1}\right|\left|{q}_{2}\right|}{{r}^{2}}\left(i\right)$

The magnitude of gravitational force $F=G\frac{Mm}{{r}^{2}}\left(ii\right)$

The number of electrons present, $n=qle\left(ii\right)$

The magnitudes of the gravitational and electrical forces must be the same. Thus, using equations (i) and (ii), we can get the value of the net positive charge as given: $(q$

is the charge on either body, $r$ is the centre - to- centre separation of Earth and Moon, $G$ is the universal gravitational constant, $M$ is the mass of Earth, and m is the mass of the Moon.)

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