Universe: Stars And Galaxies
6th Edition
ISBN: 9781319115098
Author: Roger Freedman, Robert Geller, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 1, Problem 35Q
To determine
The advantage for the astronomers to light year for measurement of distance.
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A planet's speed in orbit is given by V = (30 km/s)[(2/r)-(1/a)]0.5 where V is the planet's velocity, r is the distance in AU's from the Sun at that instant, and a is the semimajor axis of its orbit.
Calculate the Earth's velocity in its orbit (assume it is circular):
What is the velocity of Mars at a distance of 1.41 AU from the Sun?
What is the spacecraft's velocity when it is 1 AU from the Sun (after launch from the Earth)?
What additional velocity does the launch burn have to give to the spacecraft? (i.e. What is the difference between the Earth's velocity and the velocity the spacecraft needs to have?)
How fast will the spacecraft be traveling when it reaches Mars?
Does the spacecraft need to gain or lose velocity to go into the same orbit as Mars?
What would be the answers to 15-17?
Chapter 1 Solutions
Universe: Stars And Galaxies
Ch. 1 - Prob. 1QCh. 1 - Prob. 2QCh. 1 - Prob. 3QCh. 1 - Prob. 4QCh. 1 - Prob. 5QCh. 1 - Prob. 6QCh. 1 - Prob. 7QCh. 1 - Prob. 8QCh. 1 - Prob. 9QCh. 1 - Prob. 10Q
Ch. 1 - Prob. 11QCh. 1 - Prob. 12QCh. 1 - Prob. 13QCh. 1 - Prob. 14QCh. 1 - Prob. 15QCh. 1 - Prob. 16QCh. 1 - Prob. 17QCh. 1 - Prob. 18QCh. 1 - Prob. 19QCh. 1 - Prob. 20QCh. 1 - Prob. 21QCh. 1 - Prob. 22QCh. 1 - Prob. 23QCh. 1 - Prob. 24QCh. 1 - Prob. 25QCh. 1 - Prob. 26QCh. 1 - Prob. 27QCh. 1 - Prob. 28QCh. 1 - Prob. 29QCh. 1 - Prob. 30QCh. 1 - Prob. 31QCh. 1 - Prob. 32QCh. 1 - Prob. 33QCh. 1 - Prob. 34QCh. 1 - Prob. 35QCh. 1 - Prob. 36QCh. 1 - Prob. 37QCh. 1 - Prob. 38QCh. 1 - Prob. 39QCh. 1 - Prob. 40QCh. 1 - Prob. 41QCh. 1 - Prob. 42Q
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- In a part of Earth’s orbit where Earth is moving faster than usual around the Sun, would the length of the solar day change? If so, how? Explain.arrow_forwardVenus orbits 0.72 AU from the Sun. What is that distance in kilometers? (Hint: See Problem 3.)arrow_forwardIs the ecliptic the same thing as the celestial equator? Explain.arrow_forward
- thanks. Moon is at the distance 384400 km from Earth and orbits the Earth every ∼28 days. If the radius of the Moon is 1737 km (consider it to be spherical), what is the area of the moon as measured by the observer on Earth? (Hint: Length contractionarrow_forwardThe average Earth-Moon distance is 3.84 X 10^5 km, while the Earth-Sun is 1.496 X 10^8 km. Since the radius of the Moon is 1.74 X 10^3 km and that of the Sun is 6.96 X 10^5 km. a) Calculate the angular radius of the Moon and the Sun, qmax, according to the following figure. D Bax R b) Calculate the solid angle of the Moon and the Sun as seen from Earth. (c) Interpret its results; Would this be enough to explain the occurrence of total solar eclipses?arrow_forwardWhat is 1-5/6 Please tell mearrow_forward
- This was somewhat helpful, but my radius is not 3.5km it's 5km. This is not the answer to my question, it is only similar to it. I can not interpret your handwriting to follow along to answer for my question.arrow_forwardThe radius of the earth is 6400 Km and the height of a person is 1.7m, how high can you stand to see how far the horizon is from you (going back in degrees after the arc)arrow_forwardIf you find a planet's orbital period around the Sun to be ~15 years, what would be its approximate semimajor axis to the nearest AU?arrow_forward
- Suppose, hypothetically, that the Earth orbited the Sun at half its current distance. (That is, at 1/2 AU instead of 1 AU). What would be the length of the year? What else would be different?arrow_forwardAn arcminute is what fraction of a degree?arrow_forwardUse Kepler's Law, which states that the square of the time, T, required for a planet to orbit the Sun varies directly with the cube of the mean distance, a, that the planet is from the Sun.Using Earth's time of 1 year and a mean distance of 93 million miles, the equation relating T (in years) and a (in million miles) is 804375T2=a3.Use that relation equation to determine the time required for a planet with mean distance of 206 million miles to orbit the Sun. Round to 2 decimal places. yearsarrow_forward
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