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yulyashka [42]
3 years ago
11

Suppose we placed a positive charge Q on the Moon and an equal positive charge Q on the Earth. What value of Q would be needed t

o neutralize the gravitational attraction of the Moon and the Earth?
Physics
1 answer:
Rudiy273 years ago
4 0

Answer: q=5.70 x 10^13 C

Explanation:

gravitational attraction = electrostatic repulsion GMm/d^2 = kQ^2/d^2 as you can see the d^2 cancel out. that is why lunar distance is irrelevant. G is the universal gravitational constant = 6.67 x 10^-11 m^3 / kgs^2 M is earth's mass = 5.972 × 10^24 kg m is moon's mass = 7.342×10^22 kg Q is charge on earth and moon. k is coulomb's constant = 9 x10^9 N m^2 /C^2 On solving equation for Q. Q = sqrt (GMm/k) = sqrt ( 6.67 x 10^-11 x 5.972 x 10^24 * 7.342×10^22 / 9 x10^9) = 5.70 x 10^13 C

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  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

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  • We can use the same equation, to find the value of D, as follows:

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  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
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  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

3 0
3 years ago
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