Suppose that a planet were discovered between the Sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury. What would be the orbital period of such a planet? (Such a planet was once postulated, in part to explain the precession of Mercury's orbit. It was even given the name Vulcan, although we now have no evidence that it actually exists. Mercury's precession has been explained by general relativity)
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- Taking the age of Earth to be about 4 ✕ 109 years and assuming its orbital radius of 1.5 ✕ 1011 m has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).Taking the age of Earth to be about 4 ✕ 109 years and assuming its orbital radius of 1.5 ✕ 1011 m has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be 35.1km/s and 53.1km/s. The slower planet's orbital speed is 7.53years. What is the mass of the star, and what is the orbital period of the faster planet, in years?
- According to Lunar Laser Ranging experiment the average distance LM from the Earth to the Moon is approximately 3.91 x 105 km. The Moon orbits the Earth and completes one revolution relative to the stars in approximately 27.5 days (a sidereal month). Calculate the orbital velocity of the Moon in m/s. Answer: Choose...Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is V = 240 km/s and the orbital period of each is 12.1 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 x 1030 kg.) solar masses XCM MProblem 5: Suppose you are told that a satellite orbiting the Earth has a orbital period of 0.85 hours. a)Using the orbital characteristics of the Moon (RM = 3.84 × 105km and TM = 0.0748 y), use Kepler's laws to calculate the orbital radius for the satellite, in kilometers. b)What is unreasonable about this result?MultipleChoice :1) This radius is unreasonable because it is smaller than the orbital radius of the Moon.2) There is nothing unreasonable about the result.3) This radius is unreasonable because it is greater than the radius of the Earth.4) This radius is unreasonable because it is greater than the orbital radius of the Moon.5) This radius is unreasonable because it is smaller than the radius of Earth.
- Read the question carefully and give me right solution according to the question. A particle travels around a circular path having a radius of 40 m. If it is initially traveling with a speed of 10 m/s and its speed then increases at a rate of (0.05 v) m/s 2, determine the magnitude of the particle’s acceleration at t = 3 s. Round your answer to 2 decimal places.Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is v| = 225 km/s and the orbital period of each is 11.6 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 x 1030 kg.) M XCM M Part 1 of 3 - Conceptualize From the given data, it is difficult to estimate a reasonable answer to this problem without working through the details and actually solving it. A reasonable guess might be that each star has a mass equal to or slightly larger than our Sun because fourteen days is short compared to the periods of all the Sun's planets. Part 2 of 3 - Categorize The only force acting on each star is the central gravitational force of attraction which results in a centripetal acceleration. When we solve Newton's second law, we can find the unknown mass in terms of the variables…Consider the Earth's orbit around the Sun to be circular with radius R = 9.30 x 107 mi and it takes 365 days to complete one revolution. What is the distance Earth traveled for one revolution (circumference of a circle is 2??2πR )?
- We would like to be able to make meaningful interpretations of variations in the acceleration due to gravity, g, as small as 0.1 mgal. How accurate must our knowledge of the latitude of our gravity stations be? (A “gravity station" is just a location at which you make a gravity measurement.) You can assume your latitude is 45°.A satellite is in a circular orbit at an altitude of 1.5 earth radii, with an inclination of 90°. Calculate the satellite's orbital period in minutes. Report your answer to three significant digits (e.g. 123). For the radius of the earth use RE = 6371 km; for the gravitation acceleration use g = 9.81 m/s2. What is the orbital period in minutes?According to Lunar Laser Ranging experiment the average distance LM from the Earth to the Moon is approximately 3.92 x 105 km. The Moon orbits the Earth and completes one revolution relative to the stars in approximately 27.5 days (a sidereal month). Calculate the orbital velocity of the Moon in m/s. Answer: Choose...