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How do you find what the orbital period of mars in earth years?
Wiki User
∙ 12y ago
To find the answer you just Bing it. but if you want to know the answer it is 1 year, 320 days, and 18.2 hours which is nearly 1 and3/4 years.
Wiki User
∙ 12y ago
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Uranus requires 84 years to circle the sun Find Uranus' orbital radius as a multiple of Earth's orbital radius?
19.2 rE
How do you find the orbital period in earth years of a periodic comet if the furtherest from sun is 31.5 astronomical units while the closest is 0.5 astronomical units?
You can find the major axis, 0.5+31.5 or 32 AU. The semimajor axis is half that, 16 AU. Then you can use Keplers 3rd law to calculate the period, which is 161.5 or 64 years.
Venus has a period of revolution of 225 earth days Find the distance between the sun and Venus and as a multiple of Earth's orbital radius?
It takes 140 days more for the earth to go around the sun than Venus!
Why does mercury take 88 days to revolve in the sun?
This is a lot like asking: Why does the earth take 365 days to revolve around the sun ? Here's an answer, which I'm sure you'll find evasive and unsatisfying, and so it is. But when you ask "why", it really puts us out of the realm of the science, and into the philosophical realm. Kepler demonstrated, and Newton proved, that the orbital period of a light object revolving around a much more massive object under the influence of gravitation is completely determined by the dimensions of the orbit. In the case of the sun as the large central body, an average distance of 93 million miles with small eccentricity produces an orbital period of 365.25 earth days, and an average distance of 36 million miles with small eccentricity produces an orbital period of 88 earth days. 88 days is Mercury's orbital period, because 36 million miles is its distance from the sun. A grain of sand, a rock, a glob of dust, a space ship, a comet, or a planet, at the same distance from the sun, would all have an orbital period of 88 earth days.
The shape of the orbital influences the electrons?
The orbitals represent the possibility to find the electron at a particular place around the nucleus.Its an abstract term.The orbital can't affect the electron because the electron itself forms the orbital.So the orbital does not affect the electron, the electron affects the shape of the orbital.More specially, the orbital has some kind of shape because of the specific energetic condition of the electron.And with these specific, energetic conditions only specific shapes are ''allowed''.
Uranus requires 84 years to circle the sun Find Uranus' orbital radius as a multiple of Earth's orbital radius?
19.2 rE
If a planet has an average distance from the Sun of 9.1 AU, what is its orbital period SHOW YOUR WORK?
To calculate the orbital period of a planet with an average distance from the Sun of 9.1 Astronomical Units (AU), we use Kepler's Third Law of Planetary Motion: P^2 = a^3. Given: a = 9.1 AU Substitute the value of 'a' into Kepler's Third Law to find the orbital period, P: P^2 = (9.1)^3 P^2 = 753.571 AU^3 To find the orbital period P, take the square root of both sides of the equation: P = √753.571 P ≈ 27.45 years Conclusion: A planet with an average distance of 9.1 AU from the Sun has an estimated orbital period of approximately 27.45 Earth years.
How do you find the orbital period in earth years of a periodic comet if the furtherest from sun is 31.5 astronomical units while the closest is 0.5 astronomical units?
You can find the major axis, 0.5+31.5 or 32 AU. The semimajor axis is half that, 16 AU. Then you can use Keplers 3rd law to calculate the period, which is 161.5 or 64 years.
What must you know in order to find out a planets period of revolution?
Orbital information. You need to know the size of the "semi-major axis". Then you can calculate the orbital period, using Kepler's Third Law.
Venus has a period of revolution of 225 earth days Find the distance between the sun and Venus and as a multiple of Earth's orbital radius?
It takes 140 days more for the earth to go around the sun than Venus!
Why does mercury take 88 days to revolve in the sun?
This is a lot like asking: Why does the earth take 365 days to revolve around the sun ? Here's an answer, which I'm sure you'll find evasive and unsatisfying, and so it is. But when you ask "why", it really puts us out of the realm of the science, and into the philosophical realm. Kepler demonstrated, and Newton proved, that the orbital period of a light object revolving around a much more massive object under the influence of gravitation is completely determined by the dimensions of the orbit. In the case of the sun as the large central body, an average distance of 93 million miles with small eccentricity produces an orbital period of 365.25 earth days, and an average distance of 36 million miles with small eccentricity produces an orbital period of 88 earth days. 88 days is Mercury's orbital period, because 36 million miles is its distance from the sun. A grain of sand, a rock, a glob of dust, a space ship, a comet, or a planet, at the same distance from the sun, would all have an orbital period of 88 earth days.
What is Jupiters speed of travel in mph?
i am sorry i could not find the answer to your question but this might help you:A Jupiter completes an orbits every 11.86 years. This is two-fifths the orbital period of Saturn, forming a 5:2 orbital resonance between the two largest planets in the Solar System.here is another thing that might help you:Jupiter completes one orbit in 4,333 Earth days, or almost 12 Earth years.Here is the source that i looked at ( might come in handy for you)www.nasa.gov
How can you find out the distance between the Sun and Jupiter Given is Jupiter's year is 11.86 Earth years and the distance between the Sun and the Earth is 1.5x1011 meters?
Using Kepler's 3rd law of motion: T2/a3=const. T - orbital period a - distance to the Sun For Earth: T2/a3= 12/(1.5x1011)3=2.96x10-34 (11.86)2/a3=2.96x10-34 . . . a=780,095,402,531.93 meters = 5.2 AU
How do you find the period of revolution of a planet?
You have to watch it and record its position every night until it has travelled right round the ecliptic until it gets back to where you first saw it. The time taken is called the synodic period. Then you have to allow for the fact that the Earth has been going round at the same time. That gives the true orbital period.
How do you calculate that things operating in earth's gravity and in the neighborhood of earth's mass is 86 minutes?
A full derivation uses a lot of calculus, and is really too long and complicated to present here. You might find it in a college-level Physics text. Here's one way to do it for the minimum artificial-satellite period: -- One of Kepler's laws says that for every satellite in orbit around the same central body, (orbital period)2 divided by (distance)3 is the same number. You know those numbers for the moon: 27.32 days, and 238,000 miles. So if you're careful about how you handle the units, you can calculate the orbital period of any Earth satellite at any other distance from the center of the Earth. Good luck, and I'm glad that you enjoyed my answer about the mail tube, the satellite, and the pendulum.
What an orbit is and what factors affect the size speed and time period of an orbit?
An orbit is the path followed by a planet according to Kepler's laws, which are very accurate but not exact. Size, speed and period are all related by simple formulas so that if you know one you can find out the other two. The orbit is an ellipse and the size is usually measured by its mean radius, also called its semi-major axis, which is the average of the maximum and minimum distances. For the Earth that is 149.6 million kilometres. The orbital period is proportional to the mean radius to the power 1.5, while the orbital speed is inversely proportional to the square root of the mean radius.
What is the orbital notation for Sr?
if you poke yourself you will find the answer
