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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.
insert a point and into R and you'll find the velocity in m/s
You're leaving about gravitation currently correct?
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.
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.
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.