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## Homework Statement

The Earth's orbit is of mean radius [tex]1.50*10^{11}m[/tex] and the Earth's year is 365 days long.

The mean radius of orbit of Mercury is [tex]5.79*10^{10}m[/tex]. Calculate the length of Mercury's year.

## Homework Equations

[tex]F=\frac{mv^{2}}{r}[/tex]

[tex]F=\frac{Gm_{1}m_{2}}{r^{2}}[/tex]

[tex]d=v \cdot t[/tex]

## The Attempt at a Solution

[tex]v = \frac{d}{t} = \frac{2 \pi r}{t}[/tex]

[tex]\frac{mv^{2}}{r} = \frac{Gm_{1}m_{2}}{r^{2}}[/tex]

[tex]v^{2} = \frac{Gm}{r}[/tex]

subbing in [tex]v = \frac{2 \pi r}{t}[/tex] we get

[tex](\frac{2 \pi r }{t})^{2} = \frac{Gm}{r}[/tex]

by expanding;

[tex] \frac{4 \pi^{2}r^{2}}{t^{2}} = \frac{Gm}{r}[/tex]

through rearranging this we can get the formula;

[tex]\frac{r^{3}}{t^{2}} = \frac{Gm}{4 \pi^{2}}[/tex]

and since [tex]\frac{Gm}{4 \pi^{2}}[/tex] is constant for all planets (as m is the mass of the Sun) then we can assume;

[tex]\frac{R_{E}^{3}}{T_{E}^{2}} = {R_{M}^{3}}{T_{M}^{2}}[/tex]

where [tex]R_{E} , T_{E}[/tex] and [tex]R_{M} , T_{M}[/tex] are the radius of orbit and the time period of orbit of Mercury and Earth around the Sun; respectively.

Now, to find [tex]T_{M}[/tex] seems simple enough, as I am given the constants I need.

[tex]T_{M}^{2} = \frac{T_{E}^{2}}{R_{E}^{3}} \cdot R_{M}[/tex]

This gives me an answer of;

[tex]T_{M}^{2} = \frac{(365 * 86400)^{2}}{(1.50*10^{11})^{3}} \cdot (5.79*10^{10})^{3} [/tex]

I then find [tex]T_{M}[/tex] to be [tex]7.564*10^{6}s[/tex]; whereas my teacher seems to think this is wrong, and gave me 2/7 marks for this: he got his answer to 0.24...please help.