The Great Red Patch is the name given to Jupiter's large red spot. Is it possible to forecast how long it will take for the planet to complete its rotation?
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The Great Red Patch is the name given to Jupiter's large red spot.
Is it possible to forecast how long it will take for the planet to complete its rotation?

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- A projectile is fired straight upward from planet's surface with a speed that is 1/7 the planet's escape speed. Let R be the radius of the planet. What is the highest distance from the center of the planet in units of R the projectile will reach? Give your answer to three significant figures.Consider a spherical asteroid with a density p = 3000 kg m3 and radius R = 100 km. The asteroid is covered by a layer of loosely bound regolith. What is the shortest rotation period this asteroid can have without losing the regolith from its equator?Uranus has a mass of 8.68 1025 kg and a radius of 2.56 107 m. Assume it is a uniform solid sphere. The distance of Uranus from the Sun is 2.87 1012 m. (Assume Uranus completes a single rotation in 17.2 hours and orbits the Sun once every 3.06 104 Earth days.) (a) Calculate the angular momentum of Uranus in its orbit around the Sun, using kg · m2/s. (b) Calculate the angular momentum of Uranus on its axis, using kg · m2/s.
- According to Lunar Laser Ranging experiments the average distance LM from the Earth to the Moon is approximately 3.85 x 105 km. The Moon orbits the Earth and completes one revolution in approximately 27.5 days (a sidereal month). a) Calculate the orbital velocity of the Moon. b) Calculate mass of the Earth.A geosynchronous satellite is placed in orbit around Venus. Venus has a rotational period of approximately 117 earth days. How far from the surface of the Venus in km must the satellite orbit be so that it appears above the same spot of the surface?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 Moon exerts a torque on Earth's tidal bulge. Although the tides are raised in part by the Moon itself, friction between the bulge and Earth's oceans and atmosphere cause it to be in front of the Earth-Moon line in the sense of Moon's motion around Earth. Consequently , the Moon applies a torque that is acting in opposition to Earth's rotation, and that is also trying to change the direction of the axis in space. The result is ... Pick those that apply. The Moon moving farther away from Earth to conserve system total angular momentum A lengthening of the day by a second every few years The direction of Earth's axis precessing around the plane of its orbit. A speeding up of Earth's rotationa. Calculate the rotational kinetic energy (in J) of Uranus on its axis. (Assume its mass to be 8.69 ✕ 1025 kg, the period about its axis to be 17.2 h, and its diameter to be 51100 km.) b. What is the rotational kinetic energy (in J) of Uranus in its orbit around the Sun? (Assume its distance from the sun to be 2.87 ✕ 109 km and its period about the sun to be 30700 days.)34. A geosynchronous satellite appears to “hover” stationary directly above a particular point on the Earth’s equator. In order to achieve this the period of the orbit must match the sidereal period of Earth’s rotation, which is 23 hours 56 minutes. Solve for the altitude and speed of such a satellite. (Can only use Newton's 3 laws of motion and the law of gravity where Fg=G(M1M2)/r^2 and g=GM/r^2. You can also use the uniform circular motion formulas for speed and acceleration.)
- Pretend the moons orbit around the earth is a perfect circle. How long does it take in unit of days for the moon to move 90 degrees relative to the stars?The mean distance of an asteroid from the Sun is 1.74 times that of Earth from the Sun. From Kepler's law of periods, calculate the number of years required for the asteroid to make one revolution around the Sun.Q13.12 A planet (P) is moving around the sun (S) in an elliptical orbit. As the planet moves from aphelion to perihelion, the planet's angular momentum A planet P follows an elliptical orbit. The sun S is at one focus of the ellipse. Perihelion Aphelion A. increases at all times. B. decreases at all times. C. decreases during part of the motion and increases during the other part. Kea There is nothing at the other focus. D. increases, decreases, or remains the same during various parts of the motion. E. remains the same at all points between aphelion and perihelion.