Now, suppose I were to tell you that the mass of Jupiter is equal to 4.5e7 MSun. Does the stated mass of Jupiter make sense? Group of answer choices Yes No, it's too big. No, it's too small
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f the semi-major axis, a, is measured in AU and the orbital period, p, is measured in years, then Kepler's 3rd law allows us to calculate the mass of the object they are orbiting using the following equation: M = a3/p2
Furthermore, the mass that is calculated by this equation is given in solar masses (MSun) where, by definition, the Sun's mass is 1 MSun.
Now, suppose I were to tell you that the mass of Jupiter is equal to 4.5e7 MSun.
Does the stated mass of Jupiter make sense?
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- A satellite in orbit around the earth is at an altitude of 107 percent of the earth's radius. Rounding off our old friend "g" to ten meters per second squared, what is the acceleration due to earth's gravity at the satellite? Your answer should be in meters per second squared, but do not put the unit, just the number.Earth's orbit is (on average) 1 Astronomical Unit from the Sun. For reference, 1 Astronomical Unit is about 93 million miles. Jupiter's orbit is about 5 Astronomical Units from the Sun. Suppose that the gravitational force between the Earth and the Sun is equal to 50,000 N (it's way more than that but we are just pretending here). If we move Earth to Jupiter's orbit, what will be the new value of the gravitational force between the Earth and the Sun? Remember to include units of force (Newtons or N) in your answer.Please use 9.81m/s^2 for earth gravity
- Let's assume that the planet was originally orbiting around something with the mass of our Sun (and thus the mass of the planet is negligible). Now we will examine the case where the planet orbits a star with 8 times the mass of our Sun, but we will say that the dates listed in the diagram above are still accurate for indicating the planet's position as it orbits the new star. What is ratio of the size of the semimajor axis of the planet in this case compared to the original (aka how many times larger {ratio greater than 1} or smaller {ratio less than 1} is the new semimajor axis than the old one)?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°.Kepler's 1st law says that our Solar System's planets orbit in ellipses around the Sun where the closest distance to the Sun is called perihelion. Suppose I tell you that there is a planet with a perihelion distance of 2 AU and a semi-major axis of 1.5 AU. Does this make physical sense? Explain why or why not.
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- A satellite in orbit around the earth is at an altitude of 137.7 percent of the earth's radius. Rounding off our old friend "g" to ten meters per second squared, what is the acceleration due to earth's gravity at the satellite? Your answer should be in meters per second squaredUsing a diagram, define the meanings of the following terms when applied to the elliptical path of a body orbiting the Sun:(i) Semi-major axis(ii) Semi-minor axis (iii) Aphelion(iv) PerihelionState the mathematical expressions for the perihelion and the aphelion in terms of the semi-major axis, a, and the eccentricity, e.Part A Kepler's Third "Law" states that the square of the orbital period of a planet (T) is proportional to the cube of the planet's average orbital radius (r), T² = kr³, where k is the proportionality constant connecting T2 and We can derive Kepler's Third "Law" and show that k = 4/Gmgun, where G is the universal gravitational constant and msun is the mass of the sun, by %3D preventing a global thermonuclear war before it ruins our day and then binge watching the first ten seasons of the Walking Dead on Netflix. setting the acceleration due to gravity at the surface of the planet equal to the acceleration due to gravity at the surface of the sun. O setting the mutual gravitational force between the planet and the sun, Gmplanetmsunr, equal to the centripetal force acting on the planet. O stealing Robert Hooke's undeveloped idea that the mutual attractive force between the planet and the sun is proportional to 1/2. convincing ourselves that planets in circular orbits are actually in…