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Irina18 [472]
3 years ago
7

What eccentricity value results in a circular orbit?

Physics
1 answer:
Oxana [17]3 years ago
7 0

Answer:

Zero

Explanation:

Given the equation of an ellipse:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

The eccentrity of an ellipse is given by:

e=\sqrt{1-\frac{b^2}{a^2}}

For a circle, we have

a=b

Therefore the eccentricity of a circle is

e=\sqrt{1-\frac{1^2}{1^2}}=0

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Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se
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1) +2.19\mu C

The electrostatic force between two charges is given by

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where

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r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

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F= 21.63 mN = 0.02163 N

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Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

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Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

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q_2 = Q-q_1

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F=k \frac{q_1 (Q-q_1)}{r^2}

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Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

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8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

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