Use Kepler's 3rd law to find the orbital periods (assume circular orbits) for the inner planets given that their orbital radii are: Mercury: 5.8 x 107 km Venus: 1.08 x 108 km
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Q: Which unit of measurement is larger?
A: 1 rad = 57.3 degree. So for 1 rad we required 57.3 degree. So 1 degree = (1/57.3) < 1 rad
Q: If a fictitious planet has an average distance from the sun of 2.0 AU what is its orbital period
A: Using the Kepler's third law, P2=a3
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A: here is the Solution
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A: Given : Mass of Earth: 5.972×1024 kg G=6.674×10-11 m3/kg/s Period T=27.32 days
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A: Given data: The mass of the earth is m=6×1024 kg The radius is r=6.38×106 m
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A: The formula for the gravitational force is F=GmMr2
Q: Which of Kepler's law can be used to estimate the mass of a distant star system based on the period…
A: Please see the answer in step 2.
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A: Using Kepler's 3rd law, how long will it take a new planet that is 3.68 x 107 km to travel around…
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A: A) Solution is attached with image Statement of Kepler's law:
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- The solar system has a planet with an orbital period T1b=1.51d and an orbital radius of R1b=1.6456x10^6km. Another planet in the system has an orbital radius of R1f=5.5352x10^6 km. Calculate its orbital period in days.The planet Saturn orbits the Sun at an average distance of 888 million miles. Use Kepler’s 3rd law to find the period of Saturn’s orbit, in Earth years.Given the orbital radius and period of a moon, derive an algebraic solution for the mass of the planet that moon orbits. Watch those units, convert to SI.
- 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.NoneThe average radius of Earth is typically accepted to be approximately 6371 km. Using this dimension, determine the surface area in km2 and the volume in km3. Give both answers in scientific notation to one decimal place.
- Use Kepler's Law, which states that the square of the time, T, required for a planet to orbit the Sun varies directly with the cube of the mean distance, a, that the planet is from the Sun.Using Earth's time of 1 year and a mean distance of 93 million miles, the equation relating T (in years) and a (in million miles) is 804375T2=a3.Use that relation equation to determine the time required for a planet with mean distance of 206 million miles to orbit the Sun. Round to 2 decimal places. yearsOne year on Mars is as long as 1.88 years on earth. How many seconds is a Martian year? Please give the correct answerCalculate the period of a satellite orbiting the Moon, 91 km above the Moon's surface. Ignore effects of the Earth. The radius of the Moon is 1740 km. Express your answer using three significant figures and include the appropriate units. μΑ ? T = Value Units Submit Request Answer