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Vlad1618 [11]
4 years ago
12

Which planet do most known extrasolar planets most resemble?

Physics
1 answer:
tatuchka [14]4 years ago
6 0
The awnser is A>Most known exoplanets resemble gas giants known as "hot Jupiters" as a result of being large objects orbiting close to they're host star.
Most known exoplanets that we have detected are likely or confirmed to be gas giants. This is because one of the most used detection techniques is the Transit technique.

This technique involves looking at a distance star's light curve (the changes in the brightness of star over a period of time). Most stars dim and brighten over time by a small bit. But if the light curve shows a periodic and large dip in brightness, this is a sign that something large is passing (or transiting) in front of it.

The reason most detected exoplanets are gas giants is simply because they're the easiest to detect. They cause a larger dip in the curve than a smaller planet would. A small planet's dip are masked by the star's normal dimming and brightening.

Imagine having a flashlight shining on a wall. If you pass a large object through the light beam, it has a large shadow. If you pass a smaller object through, it has a smaller shadow. The change in the light you see on the wall is similar to what you will see in a star's light curve. Now if you pass an object closer to flashlight, as opposed to closer to the wall, this changes the shadow as well.

This is why most exoplanets are "hot Jupiters". They're large gas giants that are hot because they pass very close to the star, resulting in a larger dip that is easier to see in the curve. They also have shorter orbits so they're more likely to be seen in a few weeks or month's time.


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<h3>Energy of a photon</h3>

The energy of a given photon of light is calculated using the wavelength and the speed of light.

<h3>Energy of a red photon</h3>

The energy of the a red photon at the given wavelength is calculated as follows;

E_r = \frac{hc}{\lambda} \\\\E_r = \frac{6.67 \times 10^{-34} \times 3 \times 10^8}{650 \times 10^{-9}} \\\\E_r = 3.078 \times 10^{-19} \ J

<h3>Energy of an ultraviolet photon</h3>

E_r = \frac{hc}{\lambda} \\\\E_r = \frac{6.67 \times 10^{-34} \times 3 \times 10^8}{250 \times 10^{-9}} \\\\E_r = 8.0 \times 10^{-19} \ J

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Learn more about energy of a photon here: brainly.com/question/7464909

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