solve it clearly and correctly. i will rate A Martian day is 24.66 hours long. A satellite is in a circular orbit about Mars. The orbit is in the plane containing Mar's equator. Atwhat orbital radius will the satellite be in geosynchronous orbit and thereby remain directly above the same point on the equator?(MMars - 6.39 x 1023 kg, G = 6.67 x 10 11 N m2/kg2 )42,250 km22,500 km36,500 km20,400 km

Martian day is 24.66 hours long. time taken for complete revolution is T = 24.66 x 3600 sec T= 88776…

Answered: A Martian day is 24.66 hours long. A…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter11: Gravity, Planetary Orbits, And The Hydrogen Atom

Section: Chapter Questions

Problem 17P

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Since 1995, hundreds of extrasolar planets have been discovered. There is the exciting possibility that there is life on one or more of these planets. To support life similar to that on the Earth, the planet must have liquid water. For an Earth-like planet orbiting a star like the Sun, this requirement means that the planet must be within a habitable zone of 0.9 AU to 1.4 AU from the star. The semimajor axis of an extrasolar planet is inferred from its period. What range in periods corresponds to the habitable zone for an Earth-like Planet orbiting a Sun-like star?
Kepler’s third law says that the orbital period (in years) is proportional to the square root of the cube of the mean distance (in AU) from the Sun (Pa1.5) . For mean distances from 0.1 to 32 AU, calculate and plot a curve showing the expected Keplerian period. For each planet in our solar system, look up the mean distance from the Sun in AU and the orbital period in years and overplot these data on the theoretical Keplerian curve.
Calculate the effective gravitational field vector g at Earths surface at the poles and the equator. Take account of the difference in the equatorial (6378 km) and polar (6357 km) radius as well as the centrifugal force. How well does the result agree with the difference calculated with the result g = 9.780356[1 + 0.0052885 sin 2 0.0000059 sin2(2)]m/s2 where is the latitude?
Neptune orbits the Sun with an orbital radius of 4.495 x 10^12 m. If the earth to sun distance 1A.U. = 1.5 x 10^11 m, a) Determine how many A.U.'s is Neptune's orbital radius (Round to the nearest tenth). b) Given the Sun's mass is 1.99 x10^30 kg, use Newton's modified version of Kepler's formula T^2 = (4pi^2/Gm(star)) x d^3 to find the period in seconds using scientific notation. (Round to the nearest thousandth). C) Convert the period in part b) to years (Round to the nearest tenth)
Neptune orbits the Sun with an orbital radius of 4.495 x 10^12 m. If the earth to sun distance 1 A.U. = 1.5 x 10^11 m, a) Determine how many A.U.'s is Neptune's orbital radius (Round to the nearest tenth). b) Given the Sun's mass is 1.99 x 10^30 kg , use Newton's modified version of Kepler's formula T^2 = (4pi^2/Gm(star)) x d^3 to find the period in seconds using scientific notation. (Round to the nearest thousandth). c) Convert the period in part b) to years(Round to the nearest tenth).
Calculate the escape velocity to an orbit of 243 km height from a planet with the radius of 2000 km and the density of 3400 kg⋅m-³. Give your answer in Sl units. Answer: m/s ◆
A satellite is in a circular orbit 380 km above the Earth's surface. Find the orbital period and the speed of the satellite. Take MEarth = 5.97×1024 kg, REarth = 6,371 km. The orbital period, T = Units Select an answer ✓ The orbital speed, v = How high above the Earth's surface should be the orbit of a satellite so it's orbital period equals to the period of the Earth's spin (24 hrs)? The elevation above the surface, h = Units Select an answer ✓ Units Select an answer ✓
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/s
(a) Calculate Jupiter's mass given the acceleration due to gravity at the north pole is 28.328 m/s² and the radius of Jupiter at the pole is 66,850 km. kg calculated M (b) Compare this with the accepted value of 1.898 x 1027 kg. Mcalculated M. accepted
Given that the universal gravitational constant is G=6.7×10-11 N m2/kg If the radius of a planet is 6000.0km and its mass is 5.0×1024kg consider a satellite of mas 2500kg orbiting another 1000.0km above its surface and answer the questions below a) Calculate the gravitational field strength (g) from the planet on the orbiting satellite. Ignore the satellite’s radius b) What is the satellite’s ‘weight’ because of the nearby planet that it orbits around?
A sateltle orbits at a height of 6,000 km above the earth's surface. REarth = 6.38 x 10° m, mEarth = 5.98 x 1024 kg. What is this satelite's speed, in km/s (kilometer per second)? Use G = 6.67 x 10-11 N-m2/kg2. Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement. A Moving to another question will save this response.
Please answer the question and subquestions completely! This is one whole question which has subquestions! According to the official Bartleby guidelines, each question can have up to two subquestions! Thank you! 1) The weight of a 0.60 kg object at the surface of Planet V is 20 N. The radius of the planet is 4 x 10 6 m. Find the gravitational acceleration at a distance of 2 x 10 6 m from the surface of this planet. 8.9 m/s2 11 m/s2 13 m/s2 18 m/s2 A) An object accelerates if it: changes the direction of its velocity but not the magnitude changes the magnitude of the velocity but not the direction changes its speed all of the above none of the above B) A car starts from rest and in ten seconds is moving at 40 m/s. What is the car's acceleration? 0.250 m/s2 2.50 m/s2 4.00 m/s2 10.0 m/s2 40.0 m/s2 none of the above

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