- #1
Cilabitaon
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Homework Statement
The Earth's orbit is of mean radius 1.50*10^{11}m and the Earth's year is 365 days long.
The mean radius of orbit of Mercury is 5.79*10^{10}m. Calculate the length of Mercury's year.
Homework Equations
F=\frac{mv^{2}}{r}
F=\frac{Gm_{1}m_{2}}{r^{2}}
d=v \cdot t
The Attempt at a Solution
v = \frac{d}{t} = \frac{2 \pi r}{t}
\frac{mv^{2}}{r} = \frac{Gm_{1}m_{2}}{r^{2}}
v^{2} = \frac{Gm}{r}
subbing in v = \frac{2 \pi r}{t} we get
(\frac{2 \pi r }{t})^{2} = \frac{Gm}{r}
by expanding;
\frac{4 \pi^{2}r^{2}}{t^{2}} = \frac{Gm}{r}
through rearranging this we can get the formula;
\frac{r^{3}}{t^{2}} = \frac{Gm}{4 \pi^{2}}
and since \frac{Gm}{4 \pi^{2}} is constant for all planets (as m is the mass of the Sun) then we can assume;
\frac{R_{E}^{3}}{T_{E}^{2}} = {R_{M}^{3}}{T_{M}^{2}}
where R_{E} , T_{E} and R_{M} , T_{M} are the radius of orbit and the time period of orbit of Mercury and Earth around the Sun; respectively.
Now, to find T_{M} seems simple enough, as I am given the constants I need.
T_{M}^{2} = \frac{T_{E}^{2}}{R_{E}^{3}} \cdot R_{M}
This gives me an answer of;
T_{M}^{2} = \frac{(365 * 86400)^{2}}{(1.50*10^{11})^{3}} \cdot (5.79*10^{10})^{3}
I then find T_{M} to be 7.564*10^{6}s; 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.
