Question

What equal positive charges would have to be placed on two celestial objects with masses 4.69 × 1027 and 1.23 × 1025 kg to neutralize their gravitational attraction? (b) What mass of hydrogen would be needed to provide the positive charge calculated in (a)?

Answer 1

q= 2.07*10¹⁶ C

m= 2.2*10⁸ kg.

Step-by-step explanation:

a)

• In order to neutralize their gravitational atraction (given by Newton's Universal Law of Gravitation), both masses must be repelled with an equal and opposite force, that must obey Coulomb's Law.
• So, we can write the following general equation:

`G*(m_(1) *m_(2))/(r_(12)^(2)) = k * (q_(1)*q_(2))/(r_(12) ^(2)) (1)`

• Since G and k are universal constants, and that the distance on both sides is the same, and q₁ = q₂ = q, replacing in (1) the values of m₁ and m₂, we can solve for q, as follows:

`q =\sqrt{(G*m_(1)*m_(2))/(k) } = \sqrt{(6.67e-11*4.69e27*1.23e25)/(9e9) } = 2.07e16 C (2)`

b)

• Since hydrogen can carry only one elementary charge per atom, this charge will be distributed in a number of atoms that will be given by the charge q, divided by the value of the elementary charge, as follows:

`n_(H) = (q)/(e) =(2.07e16C)/(1.6e-19C) = 1.29e35 (3)`

• Since each hydrogen has one proton and one electron, neglecting the mass of the electron, the mass of hydrogen needed to provide the positive charge q, will be just the product of the number of atoms needed (given by (3)) times the mass of the proton, as follows:
• 
  `m_(H) = n_(H) * m_(p) = 1.29e35*1.67e-27 kg = 2.2e8 kg (4)`