Skip to main content
Ch 13: Gravitation

Chapter 13, Problem 13

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?

Verified Solution
Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
1452
views
Was this helpful?

Video transcript

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.
Related Practice
Textbook Question
Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 2.87 * 10^6 km from the earth and traveling at 1.20 * 10^4 km/h relative to the earth. At this time, what were (a) the spacecraft's kinetic energy relative to the earth and (b) the potential energy of the earth–spacecraft system?
1469
views
Textbook Question
On October 15, 2001, a planet was discovered orbiting around the star HD 68988. Its orbital distance was measured to be 10.5 million kilometers from the center of the star, and its orbital period was estimated at 6.3 days. What is the mass of HD 68988? Express your answer in kilograms and in terms of our sun's mass.
1230
views
Textbook Question
In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. Pluto already was known to have a large satellite Charon, orbiting at 19,600 km with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using the mass of Pluto
1007
views
Textbook Question
In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star HD 179949 (hence the term 'hot Jupiter'). The orbit was just 1 9 the distance of Mercury from our sun, and it takes the planet only 3.09 days to make one orbit (assumed to be circular). (b) How fast (in km/s) is this planet moving?
1425
views
Textbook Question
The dwarf planet Pluto has an elliptical orbit with a semimajor axis of 5.91 * 1012 m and eccentricity 0.249. (b) During Pluto's orbit around the sun, what are its closest and farthest distances from the sun?
1264
views
Textbook Question
Two uniform spheres, each with mass M and radius R, touch each other. What is the magnitude of their gravitational force of attraction?
1013
views