anonymous
  • anonymous
Find the velocity of the earth relative to the sun that we are traveling at each point in m/s
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
We'll we know its not constant correct? So it's going to change based on the gravitational field. So Fg=MeMsG/R^2 R being the distance from the sun to the earth. So if we fix this a bit, Fg=ma so Mea=MeMsG/r^2, cancel out Me's and get a=MsG/r^2 then a=v^2/r so we finally get root(MsG/r) that is the speed at a specific distance. It should be lowest at aphelion, because its the furthest away and the gravitational potential energy is highest (Ei=Ef conservation of energy here) and at perihelion, closest it should be the fast via conservation of energy again.
anonymous
  • anonymous
well*

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anonymous
  • anonymous
insert a point and into R and you'll find the velocity in m/s
anonymous
  • anonymous
what???? @awstinf
anonymous
  • anonymous
You're leaving about gravitation currently correct?
anonymous
  • anonymous
okay. so the force of gravity felt by an object orbiting the sun can be expressed as MeMsG/R^2. Where Me is the mass of earth, Ms is the mass of the sun, and G is the gravitational constant expressed as 6.67x10^-11, and R is the distanced from the sun. Force can be expressed as the force felt by the object times acceleration or F=ma. Here Fg=Me*a, where a is the acceleration felt by earth due to the force of gravity from the sun. Here this is centripetal acceleration which is expressed as v^2/R, here R is still the distance the earth is away from the sun. So using mathematical manipulations: Fg=MeMsG/R^2 Me*a=MeMsG/R^2 canacel out Me's a=MsG/R^2 Now a=v^2/R v^2/R=MsG/R^2 cancel out R's v^2=MsG/R v=Root(MsG/R) plus in a any distance R into that equation and that will give you the velocity of the earth at that point in it's orbit around the sun.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
No need of gravitation. Earth orbits the Sun in 1 year on a circle of 150 million km of radius. All you need to do is work out the length of the orbit and convert to the right units.
anonymous
  • anonymous
There isn't much physics behind doing it that way. If he's studying gravitation he should probably get a look at the topics I've listed.

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