The vernal equinox and the autumnal equinox are associated with two points 180° apart in the Earth’s orbit. That is, the Earth is on precisely opposite sides of the Sun when it passes through these two points. From the vernal equinox. 185.4 days elapse before the autumnal equinox. Only 179.8 days elapse from the autumnal equinox until the next vernal equinox. Why is the interval from the March (vernal) to the September (autumnal) equinox (which contains the summer solstice) longer than the interval from the September to the March equinox rather than being equal to that interval? Choose one of the following reasons, (a) They are really the same, but the Earth spins faster during the “summer” interval, so the days are shorter, (b) Over the “summer” interval, the Earth moves slower because it is farther from the Sun. (c) Over the March-to-September interval, the Earth moves slower because it is closer to the Sun. (d) The Earth has less kinetic energy when it is warmer, (e) The Earth has less orbital
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- When 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_forwardAccording to Lunar Laser Ranging experiments the average distance L M from the Earth to the Moon is approximately 3.85 × 105 km. The Moon orbits the Earth and completes one revolution in approximately 27.5 days (a sidereal month). Calculate mass of the Eartharrow_forwardWhat arc length does the Earth travel in a three month period in its nearly circular orbit about the sun with a radius of 1.5 x 10^(11) m?Required to answer. Single choice.arrow_forward
- The earth takes 365.256 days to go around the sun and 23 hours 56 min and 4.0905 s to revolve about its axis. These give the earth an orbital speed of 29.8 km/s and a rotational speed of 465 m/s at the equator, a 64:1 ratio. If the moon orbits the earth in a period of 29 d 12 h 44 min 2.9 s, has a sidereal period of 27.321582 days, a radius of 1737 km, and a mean distance from the earth of 384,000 km, what is the similar velocity ratio for the moon?arrow_forwardIn the year 8000 (Y8K) the mass of the Earth has increased (due to excessive space mining). As a result, the moon has moved further from the earth (the distance between the earth and the moon in the year 8000 is 4.06×108 mm ) and is now in a circular orbit. A lunar month remains at 29.5 days. Calculate the mass of the earth in the year 8000 (Y8K) M earth = ?kgarrow_forwardPlanet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs 943.0 NN on the earth weighs 920.0 NN at the north pole of Planet X and only 860.0 NN at its equator. The distance from the north pole to the equator is 18,850 kmkm, measured along the surface of Planet X. How long is the day on Planet X? Express your answer in hours. If a 45,000 kgkg satellite is placed in a circular orbit 3000 kmkm above the surface of Planet X, what will be its orbital period? Express your answer in seconds.arrow_forward
- At some point during their orbit, the location of the earth and moon relative to the sun will be as shown in the figure below. What are the magnitude in N, and direction of the forece experienced by the Moon, in degrees below the Sun-Moonline due to the Sun and the earth? (The mass of the Moon is 7.35x1022 kg, the Earth has a mass 5.97x1024, and the Sun has a mass of 2.00x1030. In the figure, the distance from the Moon to the earth is 3.84x108m and the distance from the Moon to the Sun is 1.50 x 1011marrow_forward4. A tracking station determines the following position and velocity vectors of a spacecraft in the geocentric equatorial coordinate frame: R= 10000 I + 0j+ 10000 R km V = 01+ 6.5/ + OR km/s %3D Find the satellite's COES and the flight path angle at that location.arrow_forwardA planet with mass 9.69x1023 kg orbits a star with mass 9.21x1030 kg. The orbit is circular, and the distance from the planet to the sun is 225x106 km. What is the length of a year on this planet? Give your answer in earth years (1 earth year = 31,557,600 seconds).arrow_forward
- The Sun’s center is at one focus of Earth’s orbit. How far from this focus is the other focus, (a) in meters and (b) in terms of the solar radius, 6.96 *10^8 m? The eccentricity is 0.0167, and the semimajor axis is 1.50 * 10^11 m.arrow_forwardThe time separating high tides is 12 hours and 25 minutes. Assume that the high tide occurs at 2:18 p.m. one afternoon. (a) At what time will high tide occur the next afternoon? 1x xp.m. (b) When would you expect low tides to occur the next day? X X a.m. and X X p.m.arrow_forwardThe radius of the Earth R_{E} = 6.378 * 10 ^ 6 m and the acceleration due to gravity at its surface is 9.81m / (s ^ 2) Calculate the altitude above the surface of earth in meters , at which the acceleration due to gravity is g= 1.2 m/s^2arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning