Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 11, Problem 29P
(a)
To determine
To prove:
(b)
To determine
To find: The distance between the sun and the Venus.
(c)
To determine
To Find: The length of the Venusian year.
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A saturn year is 29.5 times the ear
th year. How far is the saturn fro
m the sun if the earth is 1.50 ×
10km away from the sun?
Neptune orbits the Sun with an orbital radius of 4.495 x 10^12 m. If the earth to sun distance 1A.U. = 1.5 x 10^11 m, a) Determine how many A.U.'s is Neptune's orbital radius (Round to the nearest tenth). b) Given the Sun's mass is 1.99 x10^30 kg, use Newton's modified version of Kepler's formula T^2 = (4pi^2/Gm(star)) x d^3 to find the period in seconds using
scientific notation. (Round to the nearest thousandth). C) Convert the period in part b) to years (Round to the nearest tenth)
(a) A reasonably accurate value for the AU is 1.50 × 101 m. If the year is a × 107 s,
(2)
(a good approximation, and one easy to remember) calculate Earth's speed in km/s assuming a
circular orbit about the Sun. (b) The experimental determination (first attempted by Cavendish in
1797/98) yields G = 6.67×10-11 N-m²/kg?. Calculate the mass of the Sun. (c) On the other hand,
when we measure distances in AU and speeds in km/s, the constant in the Vis-Viva Equation is
GM = 900. Explain why this is the case. (d) Once we know the distance to the Sun, using its
angular size one could determine that the radius of the Sun is approximately Rsun = 7 × 10* m.
Calculate the ratio of the Sun's density to that of water.
Chapter 11 Solutions
Physics for Scientists and Engineers
Ch. 11 - Prob. 1PCh. 11 - Prob. 2PCh. 11 - Prob. 3PCh. 11 - Prob. 4PCh. 11 - Prob. 5PCh. 11 - Prob. 6PCh. 11 - Prob. 7PCh. 11 - Prob. 8PCh. 11 - Prob. 9PCh. 11 - Prob. 10P
Ch. 11 - Prob. 11PCh. 11 - Prob. 12PCh. 11 - Prob. 13PCh. 11 - Prob. 14PCh. 11 - Prob. 15PCh. 11 - Prob. 16PCh. 11 - Prob. 17PCh. 11 - Prob. 18PCh. 11 - Prob. 19PCh. 11 - Prob. 20PCh. 11 - Prob. 21PCh. 11 - Prob. 22PCh. 11 - Prob. 23PCh. 11 - Prob. 24PCh. 11 - Prob. 25PCh. 11 - Prob. 26PCh. 11 - Prob. 27PCh. 11 - Prob. 28PCh. 11 - Prob. 29PCh. 11 - Prob. 30PCh. 11 - Prob. 31PCh. 11 - Prob. 32PCh. 11 - Prob. 33PCh. 11 - Prob. 34PCh. 11 - Prob. 35PCh. 11 - Prob. 36PCh. 11 - Prob. 37PCh. 11 - Prob. 38PCh. 11 - Prob. 39PCh. 11 - Prob. 40PCh. 11 - Prob. 41PCh. 11 - Prob. 42PCh. 11 - Prob. 43PCh. 11 - Prob. 44PCh. 11 - Prob. 45PCh. 11 - Prob. 46PCh. 11 - Prob. 47PCh. 11 - Prob. 48PCh. 11 - Prob. 49PCh. 11 - Prob. 50PCh. 11 - Prob. 51PCh. 11 - Prob. 52PCh. 11 - Prob. 53PCh. 11 - Prob. 54PCh. 11 - Prob. 55PCh. 11 - Prob. 56PCh. 11 - Prob. 57PCh. 11 - Prob. 58PCh. 11 - Prob. 59PCh. 11 - Prob. 60PCh. 11 - Prob. 61PCh. 11 - Prob. 62PCh. 11 - Prob. 63PCh. 11 - Prob. 64PCh. 11 - Prob. 65PCh. 11 - Prob. 66PCh. 11 - Prob. 67PCh. 11 - Prob. 68PCh. 11 - Prob. 69PCh. 11 - Prob. 70PCh. 11 - Prob. 71PCh. 11 - Prob. 72PCh. 11 - Prob. 73PCh. 11 - Prob. 74PCh. 11 - Prob. 75PCh. 11 - Prob. 76PCh. 11 - Prob. 77PCh. 11 - Prob. 78PCh. 11 - Prob. 79PCh. 11 - Prob. 80PCh. 11 - Prob. 81PCh. 11 - Prob. 82PCh. 11 - Prob. 83PCh. 11 - Prob. 84PCh. 11 - Prob. 85PCh. 11 - Prob. 86PCh. 11 - Prob. 87PCh. 11 - Prob. 88PCh. 11 - Prob. 89PCh. 11 - Prob. 90PCh. 11 - Prob. 91PCh. 11 - Prob. 92PCh. 11 - Prob. 93PCh. 11 - Prob. 94PCh. 11 - Prob. 95PCh. 11 - Prob. 96PCh. 11 - Prob. 97PCh. 11 - Prob. 98PCh. 11 - Prob. 99PCh. 11 - Prob. 100PCh. 11 - Prob. 101PCh. 11 - Prob. 102PCh. 11 - Prob. 103PCh. 11 - Prob. 104PCh. 11 - Prob. 105PCh. 11 - Prob. 106PCh. 11 - Prob. 107P
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- Kepler’s third law says that the orbital period (in years) is proportional to the square root of the cube of the mean distance (in AU) from the Sun (Pa1.5) . For mean distances from 0.1 to 32 AU, calculate and plot a curve showing the expected Keplerian period. For each planet in our solar system, look up the mean distance from the Sun in AU and the orbital period in years and overplot these data on the theoretical Keplerian curve.arrow_forwardComet Halley (Fig. P11.21) approaches the Sun to within 0.570 AU, and its orbital period is 75.6 yr. (AU is the symbol for astronomical unit, where 1 AU = 1.50 1011 m is the mean EarthSun distance.) How far from the Sun will Halleys comet travel before it starts its return journey?arrow_forward(a) Calculate Venus' mass given the acceleration due to gravity at the north pole is 8.865 m/s? and the radius of Venus at the pole is 6,052 km. M. calculated | kg (b) Compare this with the accepted value of 4.868 x 1024 kg. calculated M аcсeptedarrow_forward
- (a) Calculate the orbital inclination required to place a Venus-orbiting spacecraft in a 400 km-by-900 km sun-synchronous orbit. (b) Determine the periapsis and apoapsis for a Mars-orbiting spacecraft whose orbit satisfies all the following conditions: it is sun-synchronous, its argument of perigee is constant, and its period is 7 h.arrow_forwardConsider an imaginary planet in our solar system at an average distance of25 AU from the Sun.(a) Calculate the orbital period of this planet. (b) This fictional planet has an orbital eccentricity of e = 0.4, calculatethe planet’s distance to the Sun at aphelion and perihelion. (c) Another imaginary planet in our solar system has a perihelion distanceof 12 AU from the Sun and an aphelion distance of 68 AU. Is theeccentricity of this new planet greater or less than the planet in theprevious question?arrow_forwardWhen Sedna was discovered in 2003, it was the most distant object known to orbit the Sun. Currently, it is moving toward the inner solar system. Its period is 10,500 years. Its perihelion distance is 75 AU. a. What is its semimajor axis in astronomical units? b. What is its aphelion distance?arrow_forward
- Check Your Understanding The nearly circular orbit of Saturn has an average radius of about 9.5 AU and has a period of 30 years, whereas Uranus averages about 19 AU and has a period of 84 years. Is this consistent with our results for Halley’s comet?arrow_forwardAccording to Lunar Laser Ranging experiments the average distance LM from the Earth to theMoon is approximately 3.85 × 105 km. The Moon orbits the Earth and completes one revolutionin approximately 27.5 days (a sidereal month). a) Calculate the orbital velocity of the Moon.b) Calculate mass of the Earth.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
- Pluto’s orbit around the Sun is highly elliptical compared to the planets in our Solar System. It has a perihelion distance of 29.7 AU and an aphelion distance of 49.5 AU. a) What is the semi-major axis of Pluto’s orbit, in AU? b) What is Pluto’s orbital period, in Earth years?arrow_forwardA mini satellite weighing 50 kg is going to be launched into an orbit around earth by arocket. One sidereal day (period of earth’s self-orbit) is 23 hours, 56 minutes and 4.1seconds. (a) The rocket is to be launched from a platform in the Mongolian Plateau with a heightof 3.5 km above earth surface. Calculate the acceleration due to earth’s gravity atthis elevation.(b) The satellite shall complete 10 orbits in precisely 1 sidereal day. Calculate theorbital height and orbital velocity of the satellite.(c) Once the satellite is in this orbit, what is the acceleration due to earth’s gravityexperienced by the satellite?(d) Referring to your answer in part (c), explain why the satellite does not fall towardsearth as a result of earth’s gravitational pull?(e) Calculate the total energy of the satellite in orbit, using infinity as frame ofreference.(f) Calculate the escape velocity to leave earth’s gravitational well.arrow_forward(a) Calculate Jupiter's mass given the acceleration due to gravity at the north pole is 28.328 m/s² and the radius of Jupiter at the pole is 66,850 km. kg calculated M (b) Compare this with the accepted value of 1.898 x 1027 kg. Mcalculated M. acceptedarrow_forward
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