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13 Dec 2019

Suppose that the attraction between the moon and the earth weredueto Coulomb forces rather than gravitational force. What wouldbethe magnitude of the charge required if equal but oppositechargesresided on both earth and moon? Mass of earth=5.98×1024 kg; mass of moon =7.35×1022 kg;earth-moon distance =3.84×108 m.

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Lelia Lubowitz
Lelia LubowitzLv2
17 Dec 2019
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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

the moon (m=7.36x10^22 kg) is bound to earth (m=5.98x10^25 kg) by gravity. if, instead, the force of attraction were the result of each having a charge of the same magnitude but opposite in sign, find the quantity of charge that would have to be placed on each to produce the required force.

11: How fast would you need to travel a) to escape the gravitational pull of the sun? b) As the sun begins to die, it will become a red giant. This means that its mass will remain the same but its diameter will increase substantially. When the sun becomes a red giant will its escape speed be greater than, less than, or the same as it is now?


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The average distance separating Earth and the Moon (center tocenter) is 384 000 km. Use the data in Table 7.3 to find the netgravitational force exerted by Earth and the Moon on a 3.00 104 kgspaceship located halfway between them.
1 N


Useful Planetary Data Body Mass (kg) Mean Radius (m) Period (s)Distance from Sun (m) T2/r3 (s2/m3)
Mercury 3.18 1023 2.43 106 7.60 106 5.79 1010 2.97 10-19
Venus 4.88 1024 6.06 106 1.94 107 1.08 1011 2.99 10-19
Earth 5.98 1024 6.37 106 3.156 107 1.496 1011 2.97 10-19
Mars 6.42 1023 3.37 106 5.94 107 2.28 1011 2.98 10-19
Jupiter 1.90 1027 6.99 107 3.74 108 7.78 1011 2.97 10-19
Saturn 5.68 1026 5.85 107 9.35 108 1.43 1012 2.99 10-19
Uranus 8.68 1025 2.33 107 2.64 109 2.87 1012 2.95 10-19
Neptune 1.03 1026 2.21 107 5.22 109 4.50 1012 2.99 10-19
Pluto ~1.4 1022 ~1.5 106 7.82 109 5.91 1012 2.96 10-19
Moon 7.36 1022 1.74 106 - - -
Sun 1.991 1030 6.96 108 - - -

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