Calculate the maximum and minimum possible distances between Earth and Mars (hint: you need to calculate perihelion and aphelion distance for both planets).
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Calculate the maximum and minimum possible distances between Earth and Mars (hint: you need to calculate perihelion and aphelion distance for both planets).
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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?When the Earth passes directly between the Sun and Mars, the Earth and Mars are closest to each other. If Mars is 1.52 AU from the Sun and there are 1.5 x 108 km in 1 AU, how many times will the width of the U.S. (2,530 miles) fit end-to-end between Mars and Earth? Planets and Sun not drawn to scale. Mars Earth Sun Part 1 of 4 Mars is 1.52 AU from the Sun. How many times further away from the Sun is Mars than the Earth? (The distances in AU are relative to the distance between the Sun and the Earth, so however many AU a planet is away from the Sun is how many times farther it is from Sun than Earth.) 1.52✔ 1.52 times further awayCongratulations! You just derived a version of Kepler's Third Law for Mars! Using the mass of Mars in kilograms and converting the 4.5 hours to seconds, calculate the distance from the center of the planet. GM kg 4π² ]s)² 3 = And then determine the distance (in km) from the surface. r = rm + rs rs km = km
- E Native American mascots - hor x SI Course Modules: AST 111: Intro x A Ch 21: Venus and Mars - AST 1 x © Squaring both sides and solvin x| + A webassign.net/web/Student/Assignment-Responses/submit?pos=16&dep=24621113&tags=autosave#question4793215_16 Tutorial The Magellan orbiter orbits Venus with a period of 3.26 hours. How far (in km) above the surface of the planet is it? (The mass of Venus is 4.87 x 1024 kg, and the radius of Venus is 6.05 x 103 km.) Part 1 of 3 The period of the orbiter's orbit can give us the speed at which the orbiter orbits the planet. We imagine the orbiter tracing a circle around the planet at a certain height, the speed is 2ar V = P Part 2 of 3 Next, we combine this with the circular velocity equation to determine the height above the planet's surface. GM V = 2ar GM Squaring both sides and solving for r gives the following equation. What is the exponent for r? GM Part 3 of 3 Congratulations! You just derived a version of Kepler's Third Law for Venus! Using…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/sEx/13: The mean radius of earth is 6400 km. The acceleration due to gravity at its surface is 9.8 m/s². Estimate the mass of earth. "" 11 › (G = 6.67 × 10-¹¹ Nm²/kg²)
- Calculate the surface escape velocities for Mars. rM = 3.3× 106 m, MM = 6 × 1022 kg).Calculate the escape velocity to an orbit of 214 km height from a planet with the radius of 2000 km and the density of 3400 kg·m-3Calculate the escape velocity in km/s to 2 decimal places on Mars given; • Mass of Mars = 6.24 * 1023 kg -Mm624103. • Diameter of Mars = 6794 km %3|
- 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.I. Directions: Complete the given table by finding the ratio of the planet's time of revolution to its radius. Average Radius of Orbit Times of Planet R3 T2 T?/R3 Revolution Mercury 5.7869 x 1010 7.605 x 106 Venus 1.081 x 1011 1.941 x 107 Earth 1.496 x 1011 3.156 x 107 1. What pattern do you observe in the last column of data? Which law of Kepler's does this seem to support? II. Solve the given problems. Write your solution on the space provided before each number. 1. You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth's surface. Find the following: a) Speed b) Period c) Radial Acceleration Given: Unknown: Formula: Solution: Answer: Given: Unknown: Formula: Solution: Answer: Given: Unknown: Formula: Solution: Answer:Consider an imaginary planet in our solar system at an average distance of25 AU from the Sun.(a) Calculate the orbital period of this planet. (b) This fictional planet has an orbital eccentricity of e = 0.4, calculatethe planet’s distance to the Sun at aphelion and perihelion. (c) Another imaginary planet in our solar system has a perihelion distanceof 12 AU from the Sun and an aphelion distance of 68 AU. Is theeccentricity of this new planet greater or less than the planet in theprevious question?