Suppose an asteroid has a semimajor axis of 4 AU. How long does it take the asteroid to go around the Sun? (a) 2 years (b) 4 years (c) 6 years (d) 8 years
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Suppose an asteroid has a semimajor axis of 4 AU. How long does it take the asteroid to go around the Sun? (a) 2 years (b) 4 years (c) 6 years (d) 8 years
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- A)At what altitude would a geostationary sattelite need to be above the surface of Mars? Assume the mass of Mars is 6.39 x 1023 kg, the length of a martian solar day is 24 hours 39minutes 35seconds, the length of the sidereal day is 24hours 37minutes 22seconds, and the equatorial radius is 3396 km. The answer can be calculated using Newton's verison of Kepler's third law.To model a moon in the solar system, consider a sphere with radius R and uniform mass density p. Let gm = the acceleration due to gravity on the surface of the sphere. Calculate gm for these values of R and p: R = 2.0×106 m; p= 2.7x103 kg/m^3; (in m/s^2) OA: OB: 1.509 2.007 OC: 2.669 OD: 3.549 OE: OF: 4.721 6.279 OG: 8.351 OH: 1.111x101Mars has two moons orbiting it. One moon is named Deimos ( terror/dread from Greek mythology). Deimos has a mass of 2(10)^15 kg and is 23,460 km from Mars. The mass of Mars is 6.42 (10) ^23 kg a) What is the gravitational force between Mars and Deimos? (give this answer in Scientific Notation with 2 decimal places) b) What is the velocity of Deimos as it orbits Mars? ( give this answer in decimal form to 2 decimal places) a) = FG __________________ N b) V = ________________ m/s
- In a distant star system there are many inhabitable planets. One of these planets is named Qomar. Qomar is 3.2 AU's from its star and takes 6.5 Earth years to go around its star once. There is another planet in the same star system called Ferenginar. Ferenginar is 0.9 AUs from the star. What is the length of a Ferengi year (on Ferenginar) in terms of Earth years?Planet X orbits the star Omega with a "year" that is 492 days long. Planet Y circles Omega at four times the orbital distance of planet X. How many earth days is a year on planet Y? Enter units as d.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.
- Like all planets, the planet Venus orbits the Sun in periodic motion and simultaneously spins about its axis. Just as on Earth, the time to make one complete orbit (i.e., the period of orbit) is what defines a year. And the time to make one complete revolution about its axis (i.e., the period of rotation) is what defines a day. The period of orbit for the Earth is 365.25 days and the period of rotation is 24 hours (1.00 day). But when these same values for Venus are expressed relative to Earth, it is found that Venus has a period of orbit of 225 days and a period of rotation of 243 days. So for Venus inhabitants, a day would last longer than a year! Determine the frequency of orbit and the frequency of rotation (in Hertz) on Venus.A planet with mass 8.07x1023 kg orbits a star with mass 1.18x1030 kg. The orbit is circular, and the distance from the planet to the sun is 121x106 km. What is the length of a year on this planet? Give your answer in earth years (1 earth year = 31,557,600 seconds).On the evening of an autumnal equinox day Siddhant noticed that Mars was exactly along the north-south meridian in his sky at the exact moment when the sun was setting. In other words, the Sun and Mars subtended an angle of exactly 90° as measured from the Earth. If the orbital radius of Mars is 1.52 au, What will be the approximate rise time of the mars on the next autumnal equinox day?
- If the satellite was placed in an orbit three times farther away, about how long would it take to orbit the Earth once? Answer in days, rounding to one significant figure.days Mars rotates on its axis once every 1.02 days (almost the same as Earth does). (a) Find the distance from Mars at which a satellite would remain in one spot over the Martian surface. (Use 6.42 1023 kg for the mass of Mars.)m(b) Find the speed of the satellite.m/s3. The Moon has a period of 27.3 days and a mean distance of 3.90×105 km from the center of Earth. a. Use Kepler's laws to find the period of a satellite in orbit 6.70x103 km from the center of Earth. b. How far above Earth's surface is this satellite?(a) Jupiter's third-largest natural satellite, Io, follows an orbit with a semimajor axis of 422,000 km (4.22 ✕ 105 km) and a period of 1.77 Earth days (PIo = 1.77 d). To use Kepler's Third Law, we first must convert Io's orbital semimajor axis to astronomical units. One AU equals 150 million km (1 AU = 1.50 ✕ 108 km). Convert Io's a value to AU and record the result. aIo = AU (b) One Earth year is about 365 days. Convert Io's orbital period to Earth years and record the result. PIo = yr (c) Use the Kepler's Third Law Calculator to calculate Jupiter's mass in solar units. Record the result. MJup(Io) = MSun (d) Based on this result, Jupiter's mass is about that of the Sun. Jupiter has a similar fraction of the Sun's volume. The two objects therefore have rather similar density! In fact, Jupiter has a fairly similar composition as well: most of its mass is in the form of hydrogen and helium.