Question

You are a mad scientist who wants to create a parallel universe with a gravitationnal constant G-0. In order to have planetary motion, you give a positive charge to the central star and a negative charge to the planets. If the sun has a charge of 7,580 exaCoulombs (10 18 C), what should be the charge of earth (absolute value) so that it still has a period of 365 days? Give your answer in teraCoulombs (1012 C). Use the following constants: 2 єд 8.854x10-12 C2 /N nm MSun 1.988 x 1030 kg MEarth 5.962 x 1024 kg DSun-Earth 1.496 x 10 m Note : The value for the charge of the sun is different for all students. There might be a comma in the number; it is NOT a decimal separator but rather an old fashioned thousand separator. Simply ignore it the value for the charge has no decimals. Note 2: Of course you can assume that you are able to create a mysterious force that will allow the stars and planet to stick together despite zero gravity! Your Answer

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Answer #1

For this kind of universe, the centrifugal force on Earth must be counter balanced by the Coulombic attraction force.

If Earth still has a period of 365 days then the angular velocity of the Earth can be calculated as

= 1.99 × 10-7 rad/s T 365 × 24 × 60 × 60

So, the centrifugal force will be

FmwR

Where, m = 5.962 x 1024 is mass of Earth

R = 1.496 x 1011 m is distance between Earth and Sun

So,

Fc = 5.962 × 10 24 x (1.99 × 10-)2 × 1.496 × 101-3.53 × 1022 N

Let the charge on Earth is -Q coulomb.

Given, charge on Sun (q) = 7580 x 1018 C

distance between Earth and Sun (R) = 1.496 x 1011 m

So, the electrostatic force between them will be

1 qe Fe =-

Where,

  4TTEQ

So, 7580 × (1.496 × 1011 )2 18 Fe-9×109 × 10 × Q Fe = 9 × 109

  30482.356 × 105 × Q

For a stable system ,

F_{e} = F_{c}

or,   30482356 × 105 × Q = 3.53 × 102

or,   Q = 11.587 × 1012 C = 11.6 tera Coulombs

Hence, 11.6 tera coulombs of negative charge is placed on the Earth.

For any doubt please comment and please give an up vote. Thank you.

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