The star Rho1 Cancri is 57 light-years from the earth and has a m...

Question

The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho1 Cancri with an orbital radius equal to 0.11 times the radius of the earth’s orbit around the sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho1 Cancri?

Accepted Answer

Welcome back everybody. We are looking at the planet kepler 186 f more colloquially known as Earth's cousin. And it is orbiting a red dwarf star. It is orbiting this direction. And we are told a couple different things. We're told that the distance between the two bodies is 52.4 million kilometers or 52.4 times 10 to the ninth meters. Now we are also told that the mass of kepler, our cousin is 0. times the mass of the sun. This is equivalent to 0.48 times 1.99 times 10 to the 30th kilograms, which just works out that the mass of kepler F is 9.552 times to the 29th kilograms. Wonderful. And we are asked to find two different things here. We are asked to find what the orbital speed of kepler is and what its orbital period is. Now, kepler, the individual who the planet is named after. Give us some pretty helpful formulas for this. So let's go ahead and start with the orbital speed here, we're told that the orbital speed of a planet is equal to the square root of Newton's gravitational constant, times the mass of the orbiting body. Our planet in this case, all over the distance between the orbiting body and the body it orbits. Let's go ahead and plug in some values here, gravitational constant is given by 6. times 10 to the negative 11th. This is going to be times the mass of our planet which we determined to be 9.552 times to the 29th. All over the distance in meters between the two, which is going to be 52.4 times 10 to the ninth, plugging all of this into our calculator. We get that this is 2.6 times 10 to the fourth meters per second. Wonderful. Now that we have the orbital speed, we can use that to determine the orbital period. We are given that the orbital period is equal to two pi times the distance between the two bodies, all divided by the orbital speed. So let's once again just plug in all the values that we know. We have two pi times the distance between the two, which is 52.4 times 10 to the ninth. All divided by the orbital speed that we just found of 2.6 times 10 to the fourth. When you plug this into your calculator, you get that this is equal to 1.59 times 10 to the seven seconds. But we want this in days. So I'm gonna divide this by 60 since there are 60 seconds in a minute, times 60 since there are 60 minutes in an hour, Times 24 since there are 24 hours in a day, giving us 185 days. We have now found the orbital speed in the orbital period corresponding to answer choice. B Thank you all so much for watching hope. This video helped. We will see you all in the next one.

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