Hot Jupiters. In 2004 astronomers reported the discovery of...

Hot Jupiters. 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 $\frac{1}{9}$ the distance of Mercury from our sun, and it takes the planet only 3.09 days to make one orbit (assumed to be circular). (a) What is the mass of the star? Express your answer in kilograms and as a multiple of our sun's mass. (b) How fast (in $\mathrm{km} / \mathrm{s}$ ) is this planet moving?

Key Concepts

Kepler's Third Law

Kepler's Third Law relates the orbital period of a body to its orbital radius and the mass of the central object. In astrophysics, it is used to determine either the mass of a star or the characteristics of an orbit when the other parameters are known. This law underpins many calculations in celestial mechanics and provides a straightforward method for estimating unknown quantities by comparing to familiar benchmark systems.

Newton's Law of Universal Gravitation

This law describes the attractive force between two masses and is fundamental in analyzing the motion of celestial bodies. It states that the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. This principle is essential for understanding how planets orbit stars and for balancing gravitational forces with the centripetal force required for circular motion.

Centripetal Force in Circular Motion

For an object moving in a circular orbit, the required centripetal force to maintain that motion must be provided by gravity. By equating the gravitational force to the centripetal force, one can derive relationships between orbital speed, radius, and the mass of the central body. This concept is vital for determining orbital velocities and understanding the dynamics of orbiting systems.

Unit Conversion and Dimensional Analysis

In astrophysical calculations, it is crucial to consistently manage units across various physical quantities, such as converting distances and times into standard units like meters and seconds. This process, known as unit conversion and dimensional analysis, ensures that the equations used in orbital mechanics yield accurate and meaningful results, allowing comparisons between different astrophysical objects and systems.

Transcript

00:01 So in this question, we are talking about hot jupiters.

00:04 So in 2004, there was a discovery of a large jupiter -sized planet orbiting a star very, very close to it.

00:14 That's why we call them hot jupiters because they are very close to their stars, and hence they are very hot.

00:22 The orbit is just one -ninth, the distance of mercury from our sun, and it takes the planet only 3 .090s to make the orbit.

00:31 Assuming it's circular.

00:34 We want to calculate what the mass of the star is and how fast the planet is moving in kilometers per second.

00:41 So first of all, let's parse out some of the information that was given in the question.

00:47 So we've been told that the orbital radius of the planet is one ninth, the distance between mercury and our sun.

00:58 So if you look at appendix f in the planet, book, you'll see that that distance is 5 .79 times 10 to the 10 meters.

01:08 And so we can calculate the orbital radius of this new planet by dividing by 9 or multiplying by 1 9th, and we get 6 .43 times 10 to the 9 meters.

01:19 So that's what we're going to be using for our orbital radius here.

01:24 And then the orbital period was given in terms of days.

01:28 And for formulas that we use, we normally, want to have this in seconds.

01:33 So we're going to multiply by 24 hours in a day and 3 ,600 seconds per hour, and that will give us the period in seconds, 2 .67 times 10 to the 5 seconds.

01:56 So now that we have this, we can go ahead and calculate the mass of the star.

02:05 Now, how are we going to do that? well, what we can do is we can kind of reverse engineer the formula for orbital speed.

02:15 And so i can rearrange this to get m, capital m by itself.

02:20 So i'm going to square both sides.

02:23 I'm going to multiply by r0, and it's divided by g on both sides.

02:30 And so that'll give a formula for m of v squared times r0.

02:37 Over g.

02:40 So this is the formula we're going to use.

02:42 The only problem is that we don't have what v is.

02:45 So we're going to go ahead and calculate v very quickly on the side here using the orbital radius and the period.

02:56 So we're just basically using distance over time here where the distance traveled by the planet is the circumference of the circle to pi r0...

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