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Q18.8-68PE

Expert-verifiedFound in: Page 668

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible compared to the gravitational attraction between the ships. Construct a problem in which you place identical excess charges on the space ships to exactly counter their gravitational attraction. Calculate the amount of excess charge needed. Examine whether that charge depends on the distance between the centers of the ships, the masses of the ships, or any other factors. Discuss whether this would be an easy, difficult, or even impossible thing to do in practice.**

(a) In order to counteract the gravitational attraction a ship must have\(4.3 \times {10^{ - 5}}{\rm{ C}}\) charge on them. (b) No, the charge on the ship is unaffected by the distance between the ships' centres.

**According to the universal law of gravitational, every object attracts another object towards its center with a force known as gravitational force.** The gravitational force exists due to the mass of the object.

The two identical spaceships of mass \(5 \times {10^5}{\rm{ kg}}\) are \(50{\rm{ m}}\) from each other. (a) How much charge should each ship carry to counteract the gravitational attraction between them? (b) Is the charge on the ship affected by the distance between the ships' centres?

The gravitational force of attraction between two identical space ships of mass \(m\) and separated by a distance \(r\) is,

\({F_g} = \frac{{G{m^2}}}{{{r^2}}}\)

Here, \(G\) is the universal gravitational constant.

The electrostatic force between two identical space ships having same magnitude of charge \(q\) and separated by a distance \(r\) is,

\({F_e} = \frac{{K{q^2}}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant.

Since, the gravitational force of attraction is balanced by the electrostatic force. Therefore,

\(\begin{array}{c}{F_e} = {F_g}\\\frac{{K{q^2}}}{{{r^2}}} = \frac{{G{m^2}}}{{{r^2}}}\end{array}\)

Rearranging the above expression in order to get an expression for the charge,

\(q = \sqrt {\frac{{G{m^2}}}{K}} \ldots \left( {1.1} \right)\)

Substitute \(6.67 \times {10^{ - 11}}{\rm{ N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}\) for \(G\), \(5 \times {10^5}{\rm{ kg}}\) for \(m\), and \(9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\) for\(K\),

\(\begin{array}{c}q = \sqrt {\frac{{\left( {6.67 \times {{10}^{ - 11}}{\rm{ N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}} \right) \times {{\left( {5 \times {{10}^5}{\rm{ kg}}} \right)}^2}}}{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)}}} \\ = 4.3 \times {10^{ - 5}}{\rm{ C}}\end{array}\)

Hence, the charge on the ship is \(4.3 \times {10^{ - 5}}{\rm{ C}}\).

From equation \(\left( {1.1} \right)\), it is clear that the charge is independent of the distance between the centers of space ship.

Hence, the charge on the ship does not depend on the distance between the centers of the ships.

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