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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.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?The earth revolves around the sun in exactly 365 1/4 days which is equivalent to 1 year. To make up for the loss of 1/4 day, the calendar was adjusted so that we have a leap year for every 4 years. If the earth were to speed in its motion slightly so that a year would be completed in exactly 365 days and 6 hours, how often would we need to have a leap year?
- The time it takes the Earth to complete one 360-day rotation around its axis is 24 hours, 4 minutes 24 hours 23 hours, 56 minutesIt takes the Earth approximately 365 normal days to orbit the sun. Using this information and assuming 24 hours in a normal day, approximately how many nanoseconds does it take for the Earth to orbit the sun?What would be the duration of the year if the distance between the earth and the sun gets tripled.
- I measured the angular separation of Jupiter's moons in arcminutes/arcseconds and converted this angle to radians. The conversion factor for degrees to radians was 57.3 degrees per radian. Which unit of measurement is larger? 1 degree or 1 radian?The planet Earth has a semi-major axis of a = 1.00 AU and an orbital period of P= 1 sidereal year = 365.25 days = 3.156 x 10^7 s. Compute the orbital periods of bodies orbiting the Sun with each of the following semi-major axes. a) a = 0.1 AU b) a = 10 AU c) a = 100 AU d) a = 1000 AU e) a = 10,000 AU 1 AU = 1.496 x 10^8 km = 1.496 x 10^11 m = 1.496 x 10^13 cm. GM(sun) = 1.327 x 10^20 m^3/s^2 = (Newton's Constant) x (Mass of Sun) %3D %3DWhat should be the duration of the year if the distance between the earth and the sun gets doubled?
- The solar system has a planet with an orbital period T1b=1.51d and an orbital radius of R1b=1.6456x10^6km. Another planet in the system has an orbital radius of R1f=5.5352x10^6 km. Calculate its orbital period in days.The dwarf planet Praamzius is estimated to have a diameter of about 300 km and orbits the sun at a distance of 6.8×10^12m. What is its orbital period in years? Express your answer in years to three significant figures.If G = 6.674 ⨉ 10 −11m3/kg/s 2and M Earth= 5.972 ⨉ 10 24kg and the sidereal period of the Earth is 27.32 days, then, from Kepler’s third law in #4, what is the expected orbital distance of the Moon?