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Alexus [3.1K]
3 years ago
13

An exoplanetary system has two known planets. Planet X orbits in 290 days and Planet Y orbits in 145 days. Which planet is close

st to its host star? If the star has the same mass as the Sun, what is the semi-major axis of the orbits for Planets X and Y?
Physics
1 answer:
satela [25.4K]3 years ago
6 0

Answer:

Planet Y

Rx = 1.587 Ry

Explanation:

Tx = 290 days

Ty = 145 days

Let the semi major axis of planet X is Rx and of plant Y is Ry.

According to the Kepler's third law of planetary motion

\frac{T_{x}^{2}}{T_{y}^{2}}=\frac{R_{x}^{3}}{R_{y}^{3}}

\frac{290^{2}}{145^{2}}=\frac{R_{x}^{3}}{R_{y}^{3}}

4=\frac{R_{x}^{3}}{R_{y}^{3}}

1.587=\frac{R_{x}}{R_{y}}

Rx = 1.587 Ry

So, planet Y is closest to star.

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To develop this problem we require the concepts related to Frequency and their respective way of calculating it.

The formula to calculate the frequency is given by

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Considering the Coulomb's Law, the magnitude of the Coulomb force is 3.1865 N.

<h3>Coulomb's Law</h3>

Charged bodies experience a force of attraction or repulsion on approach.

From Coulomb's Law it is possible to predict what the electrostatic force of attraction or repulsion between two particles will be according to their electric charge and the distance between them.

From Coulomb's Law, the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

F=k\frac{Qq}{d^{2} }

where:

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  • Q and q are the values ​​of the two point charges. They are measured in Coulombs (C).
  • d is the value of the distance that separates them. It is measured in meters (m).
  • K is a constant of proportionality called the Coulomb's law constant. It depends on the medium in which the charges are located. Specifically for vacuum k is approximately 9×10⁹ \frac{Nm^{2} }{C^{2} }.

The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.

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In this case, you know that:

  • The two uncharged sphere are separated by the distance of d= 3.50 m
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Replacing in Coulomb's Law:

F=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{(2.0826x10^{-7} C)x(2.0826x10^{-7} C)}{(3.50 m)^{2} }

Solving:

<u><em>F= 3.1865 N</em></u>

Finally, the magnitude of the Coulomb force is 3.1865 N.

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