21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 10, Problem 34QP
To determine
The estimation of the radius of Io’s plasma torus in terms of the Jupiter’s radius using figure 10.17 and verify it with answer on the internet.
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Calculate the escape velocity to an orbit of 243 km height from a planet with the radius of 2000 km and the density of 3400 kg m-³. Give your answer in
Sl units.
Answer:
Choose...
You can determine the radius of a planet by measuring the change in the flux coming from the star (i.e. the transit method). If the radius of the planet increases, the overall flux decreases, but what happens to the change in the flux?
Hint: the change in a quantity is represented by the uppercase delta or triangle symbol.
Group of answer choices
- Increases
- Decreases
- Stays constant
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Tutorial
Based on the orbital properties of Uranus, how far across the sky in arc seconds does it travel in one Earth
day? The average orbital radius is 2.88 x 109 km and the period is 84.0 years. (Assume Uranus and the Earth
are at the closest point to one another in their orbits.)
How many full Moons does this distance cover if the Moon has an angular diameter of 0.5 degrees?
Part 1 of 4
We first need to determine how fast the planet is moving across the sky. If we know the period and the
distance between the Sun and the planet we can calculate the velocity using:
2ar
which will tell us how many kilometers the planet travels in a day if we convert the period into days.
days
= (P
years'
|days/year
Pdays
days
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Chapter 10 Solutions
21st Century Astronomy
Ch. 10.1 - Prob. 10.1CYUCh. 10.2 - Prob. 10.2CYUCh. 10.3 - Prob. 10.3CYUCh. 10.4 - Prob. 10.4CYUCh. 10.5 - Prob. 10.5CYUCh. 10 - Prob. 1QPCh. 10 - Prob. 2QPCh. 10 - Prob. 3QPCh. 10 - Prob. 4QPCh. 10 - Prob. 5QP
Ch. 10 - Prob. 6QPCh. 10 - Prob. 7QPCh. 10 - Prob. 8QPCh. 10 - Prob. 9QPCh. 10 - Prob. 10QPCh. 10 - Prob. 11QPCh. 10 - Prob. 12QPCh. 10 - Prob. 13QPCh. 10 - Prob. 14QPCh. 10 - Prob. 15QPCh. 10 - Prob. 16QPCh. 10 - Prob. 17QPCh. 10 - Prob. 18QPCh. 10 - Prob. 19QPCh. 10 - Prob. 20QPCh. 10 - Prob. 21QPCh. 10 - Prob. 22QPCh. 10 - Prob. 23QPCh. 10 - Prob. 24QPCh. 10 - Prob. 25QPCh. 10 - Prob. 26QPCh. 10 - Prob. 27QPCh. 10 - Prob. 28QPCh. 10 - Prob. 29QPCh. 10 - Prob. 30QPCh. 10 - Prob. 31QPCh. 10 - Prob. 32QPCh. 10 - Prob. 33QPCh. 10 - Prob. 34QPCh. 10 - Prob. 35QPCh. 10 - Prob. 36QPCh. 10 - Prob. 37QPCh. 10 - Prob. 38QPCh. 10 - Prob. 39QPCh. 10 - Prob. 40QPCh. 10 - Prob. 41QPCh. 10 - Prob. 42QPCh. 10 - Prob. 43QPCh. 10 - Prob. 44QPCh. 10 - Prob. 45QP
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- At what average distances (in AU) from the Sun would you expect to find Kirkwood gaps where the orbital period of asteroids are respectively four-ninths, and two-fifths, of the orbital period of Jupiter? (Hint: use Kepler's third law.) inner gap at AU outer gap at AU Compare your results with the figure. 7:2 3:1 8,3 5;2 7:3 9:4 2;1 9:5 7:4 53 8:5 3:2 Resonances Gap Gap Gap Gap GapGap Gap Distance from sun (AU) O The results agree with the figure. O The results do not agree with the figure.arrow_forwardCalculate the gravitational field strength on a planet with a diameter of 15487.3 km and the escape velocity of 11 km s-1. Give your answer in SI units.arrow_forwardThe value we have just calculated is the combined masses of Jupiter and Callisto! Their mass is in units of the Sun's Mass (MS) - we can convert this to units which are more familiar to us like kilograms by multiplying this answer by the scale factor (1.99x1030 kg/1 MS): (MJupiter + MCallisto) = ( MS) (1.99x1030 kg/1 Solar Mass) = _______x_10___ kg (I have already written the x 10 so you are reminded to write the exponenet of 10 in the scientific notation expression of your answer). Since you know from looking at pictures of Jupiter with its Galilean Satellites (look in your book at a picture if you have not already), that Callisto is much smaller than Jupiter - in fact it is less than 0.001 of Jupiter's size or mass, so the number we have just calculated for (MJupiter + MCallisto) is almost the same as MJupiter . How much more massive is Jupiter than the Earth? (The mass of Earth is about 5.98 x 1024 kg)arrow_forward
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