Stars and Galaxies (MindTap Course List)
10th Edition
ISBN: 9781337399944
Author: Michael A. Seeds
Publisher: Cengage Learning
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Chapter 4, Problem 21RQ
To determine
The value of
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Kepler's 1st law says that our Solar System's planets orbit in ellipses around the Sun where the closest distance to the Sun is called perihelion.
Suppose I tell you that there is a planet with a perihelion distance of 2 AU and a semi-major axis of 1.5 AU.
Does this make physical sense? Explain why or why not.
I. Directions: Complete the given table by finding the ratio of the planet's time of revolution to its radius.
Average
Radius of
Orbit
Times of
Planet
R3
T2
T?/R3
Revolution
Mercury
5.7869 x 1010
7.605 x 106
Venus
1.081 x 1011
1.941 x 107
Earth
1.496 x 1011
3.156 x 107
1. What pattern do you observe in the last column of data? Which law of Kepler's does this seem to support?
II. Solve the given problems. Write your solution on the space provided before each number.
1. You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth's surface. Find the
following:
a) Speed
b) Period
c) Radial Acceleration
Given:
Unknown:
Formula:
Solution:
Answer:
Given:
Unknown:
Formula:
Solution:
Answer:
Given:
Unknown:
Formula:
Solution:
Answer:
Show your complete and detailed solution.
Round off your answers to 4 decimal digits and box your final answers.
Chapter 4 Solutions
Stars and Galaxies (MindTap Course List)
Ch. 4 - Prob. 1RQCh. 4 - Prob. 2RQCh. 4 - Prob. 3RQCh. 4 - Prob. 4RQCh. 4 - Prob. 5RQCh. 4 - Prob. 6RQCh. 4 - Which two-dimensional (2D) and three-dimensional...Ch. 4 - Prob. 8RQCh. 4 - Prob. 9RQCh. 4 - Prob. 10RQ
Ch. 4 - Prob. 11RQCh. 4 - Prob. 12RQCh. 4 - Prob. 13RQCh. 4 - Prob. 14RQCh. 4 - Assume the night is clear and the Moons phase is...Ch. 4 - Prob. 16RQCh. 4 - Prob. 17RQCh. 4 - Prob. 18RQCh. 4 - Prob. 19RQCh. 4 - Prob. 20RQCh. 4 - Prob. 21RQCh. 4 - Prob. 22RQCh. 4 - How did the Alfonsine Tables, the Prutenic Tables,...Ch. 4 - Prob. 24RQCh. 4 - Prob. 25RQCh. 4 - Prob. 26RQCh. 4 - Prob. 27RQCh. 4 - Draw and label a diagram of the western horizon...Ch. 4 - Prob. 2PCh. 4 - Prob. 3PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Prob. 7PCh. 4 - One planet is three times farther from the Sun...Ch. 4 - Prob. 9PCh. 4 - Prob. 10PCh. 4 - Prob. 11PCh. 4 - Prob. 1SPCh. 4 - Prob. 2SPCh. 4 - Prob. 1LLCh. 4 - Prob. 2LLCh. 4 - What three astronomical objects are represented...Ch. 4 - Prob. 4LL
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- According to Kepler’s second law, where in a planet’s orbit would it be moving fastest? Where would it be moving slowest?arrow_forwardSuppose a planet sweeps out an area, call it A, in one day. In terms of A, Kepler's second law tells us it would sweep out how large an area in five days?arrow_forwardThe table below presents the semi-major axis (a) and Actual orbital period for all of the major planets in the solar system. Cube for each planet the semi-major axis in Astronomical Units. Then take the square root of this number to get the Calculated orbital period of each planet. Fill in the final row of data for each planet. Table of Data for Kepler’s Third Law: Table of Data for Kepler’s Third Law: Planet aau = Semi-Major Axis (AU) Actual Planet Calculated Planet Period (Yr) Period (Yr) __________ ______________________ ___________ ________________ Mercury 0.39 0.24 Venus 0.72 0.62 Earth 1.00 1.00 Mars 1.52 1.88 Jupiter…arrow_forward
- In your own words, describe the meaning of Kepler's Third Law of Planetary Motion. Do not use any equations, do not describe the equations in words, just tell me the conceptual meaning.arrow_forwardWhat is planetary motion through Kepler's law?arrow_forwardWrite down an expression for the gravitational filed strength of a planet of radius R and density ρ. Please use "*" for products (e.g. B*A), "/" for ratios (e.g. B/A) and the usual "+" and "-" signs as appropriate without the quotes). For Greek letters such as ?ρ and ?π use rho and pi. Please use the "Display response" button to check you entered the answer you expectarrow_forward
- The moon has a period of 27.3 days and a mean distance of 3.9 x 10 ^5 km from the center of the earth. a) Use Kepler's laws to find the period of a satellite in orbit 6.7x 10^3 km from center of earth. ( Round your answer to the nearest thousandth of a day) b) Convert your answer to minutes. (Round your answer to the nearest tenth). c) Given the earth's radius is 6.4 × 10^6 m, determine how far above earth's surface is the satellite in meters, km and miles.arrow_forwardThe moon has a period of 27.3 days and a mean distance of 3.9 x 10 ^5 km from the center of the earth. a) Use Kepler's laws to find the period of a satellite in orbit 6.7 x 10^3 km from center of earth. ( Round your answer to the nearest thousandth of a day) b) Convert your answer to minutes. ( Round your answer to the nearest tenth). c) Given the earth's radius is 6.4 x 10^6 m, determine how far above earth's surface is the satellite in meters, km and miles.arrow_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_forward
- Use 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_forwardThe angle on the sky between Venus and the Sun is measured to be 46.3° when Venus is at greatest eastern elongation. What is the distance of Venus from the Sun, measured in AU? Choose the answer below that most closely matches your answer. Select one: а. 1.763 AU O b. 0.587 AU Ос. 0.652 AU O d. 0.846 AU Ое. 0.723 AUarrow_forwardBACKGROUND An ingenious solution to the Earth's circumference occured in 230 BC. Eratosthenes, a Greek geographer, mathematician, music theorist, poet, astronomer, and philosopher, was reading in the Library of Alexandria when he noticed an account for a deep well near Syene (now Aswan), some distance to the south (800 km) in which at high noon on the longest day of the year the bottom of the well was fully illuminated by the Sun. Eratosthenes exclaimed "Ah-ah!" (or something like that), "I can solve for the circumference of the Earth!". In his mind's eye, Eratosthenes could see that at Syene, at the moment when the bottom of the well was fully lit, the Sun must have been at the Zenith (directly overhead). Yet he knew that at the same moment in Alexandria vertical objects (like a tower, pole) cast shadows. Here is the experiment perfomed by Eratosthenes (see the picture below). • He erected a vertical pole at Alexandria (A) and measured the angle of its shadow at the moment when the…arrow_forward
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