Find the velocity of the earth relative to the sun that we are traveling at each point in m/s

Question

anonymous · Answer

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anonymous · Answer

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 · Answer

well*

anonymous · Answer

insert a point and into R and you'll find the velocity in m/s

anonymous · Answer

what???? @awstinf

anonymous · Answer

You're leaving about gravitation currently correct?

anonymous · Answer

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 · Answer

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 · Answer

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.