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)?

Short Answer

Expert verified
  • a)The value of equal positive charges to be placed ontheEarth and on the Moon to neutralize their gravitational attraction is5.7×1013C .
  • b)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.
  • c)The amount of hydrogen ions needed to provide the positive charge calculated in (a) is6.0×105kg .

Step by step solution

01

Step-by-Step Solution

Positive charges are needed to be placed ontheEarth and on theMoonto neutralize their gravitational attraction.

02

Understanding concept of gravitational force and electric force

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 ofelectrostatic force between any two particles,F=kq1q2r2. (i)

The magnitude of gravitational forceF=GMmr2,. (ii)

The number of electrons present,n=q/e. (ii)

03

a) Calculation of positive charge

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 follows:kq2r2=GMnr2(sinetheelectroststicforceisbeteentwosimilarcharges)q=4πε0GMn
where 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.



Now,onsubstituting the given values in the above equation,we get

q=6.67×10-11Nm2/kg27.36×1022kg5.98×1024kg9.00×109Nm2/C2=5.7×1013C

Hence, the value of the positive charges is 5.7×1013C.

04

b) Stating reason for not taking lunar distance into consideration

From the calculations of part (a), we can see that the distance r cancels because both the electric and gravitational forces are proportional to 1/r2. Thus, the value of lunar distance is not considered.

05

c) Calculation of total mass of hydrogen ions

The charge on a hydrogen ion

e=1.60×10-19C

Now, using the charge value from part (a) in equation (iii),the total number of hydrogen ionsis found to be

n=5.7×1013C1.60×10-19C=3.6×1032ions

Each positive ion has a mass ofm1=1.67×10-27kg, so that the total mass needed

m=nml=3.6×10321.67×10-27kg=6.0×105kg

Hence, the total mass of the hydrogen ions needed is 6.0×105kg.

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