Calculate the period of revolution of planet Jupiter round the sun, given that the ratio of the radius of Jupiter's orbit to that of the earth's orbit is 5.2.
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![Calculate the period of revolution of
planet Jupiter round the sun, given that the
ratio of the radius of Jupiter's orbit to that of
the earth's orbit is 5.2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F929be09e-f184-40c2-a9b6-fc5424141446%2F872233c3-3580-4e23-be42-8b1a3f491598%2Fa7yidm9_processed.jpeg&w=3840&q=75)
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- Jupiter has lots of moons in orbit around it. One of those moons, Io, was first observed by Galileo (using an early telescope). Io has an average orbital radius about Jupiter of about 4.2 x 108 m and an orbital period of about 1.8 days. Using this data, calculate Jupiter’s mass.Scientists have discovered a distant planet with a mass of 8.2x1023 kg. The planet has a small moon that orbits with a period of 6 hours and 36 minutes. Use only this information (and the value of G) to calculate the radius of the moon's orbit (in units of 106 m).A satellite orbits the Earth at a radius of 2.2×10 7 m. What is its orbital period?
- Calculate the mass of the planet (M) by applying Kepler’s 3rd law of the period (T) and radius (r) of a circular for circular orbits at a height of (a) 2000Km and (b) 6000 km . Compare these values. Are they same or different?The orbit of a 1.5 x 1010 kg comet around the Sun is elliptical, with an aphelion distance of 11.0 AU and perihelion distance of 0.890 AU. (Note: 1 AU = one astronomical unit = the average distance from the Sun to the Earth = 1.496 x 101" m.) (a) What is its orbital eccentricity? (b) What is its period? (Enter your answer in yr.) yr (c) At aphelion what is the potential energy (in J) of the comet-Sun system?The tidal acceleration due to the moon at the surface of the earth is approximately 2.44*10^-5 m/s^2. If the moon was closer to the earth by a factor of 0.9. What then would the tidal acceleration be?
- Based on Kepler's laws and information on the orbital characteristics of the Moon, calculate the orbital radius for an Earth satellite having a period of 1.00h. What is unreasonable about this result?(a) Suppose that your measured weight at the equator is one-half your measured weight at the pole on a planet whose mass and diameter are equal to those of Earth. What is the rotational period of the planet? (b) Would you need to take the shape of this planet into account?An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 1810 × 103 seconds (about 21 days) on average to complete one nearly circular revolution around the unnamed planet. If the distance from the center of the moon to the surface of the planet is 255.0 × 106 m and the planet has a radius of 3.30 × 106 m, calculate the moon's radial acceleration ?cac.
- An artificial satellite is in a circular orbit 5.50×102 km from the surface of a planet of radius 4.50×103 km. The period of revolution of the satellite around the planet is 4.00 hours. What is the average density ?avg of the planet?An astronomer observes that a small moon is in circular orbit around a distant planet. If the moon has a period of 150.0 hours and an orbital radius of 3.70 x 10^7 m, what is the mass of the planet? Submit your answer in units of 10^23 kg.