Gravitational Fields and Satellite Motion

[Cilabitaon]

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.

 

[kuruman]

He got his answer to be 0.24 what? To see if you are right, look up the orbital period of Mercury (usually given in days) and convert to seconds. It's as easy as that.

 

[Cilabitaon]

He got his answer to be 0.24 what? To see if you are right, look up the orbital period of Mercury (usually given in days) and convert to seconds. It's as easy as that.

Yep, he gets his answer to be 0.24, and no matter what I try I can't seem to get his answer, every conversion just fails for me =[

 

[kuruman]

Can you please tell me, what the units of this 0.24 are?

 

[Cilabitaon]

Can you please tell me, what the units of this 0.24 are?

That's the point, he's so anal about not doing any work for us; he always says "In order to understand the problem you must be utterly baffled beforehand"; so he just says "you should have 0.240" and leaves it at that.

edit:

I am going to kill both myself and my Prof.

The units are years.

$$\frac{7.564 \times 10^{6}}{(3600 \times 24 \times 365)} = 0.23985 Earth years$$

 

[kuruman]

I understand your frustration and anger and I think that both feelings are not conducive to learning. However, I think that your teacher is trying to "baffle" you in an inappropriate manner. In my book, any number should be followed by the appropriate units. Conversely, if I ask someone to find a number, if the number is correct and the units are correct, then the answer is correct. I wish you fortitude, humor the guy and this too shall pass.

 

[Cilabitaon]

I understand your frustration and anger and I think that both feelings are not conducive to learning. However, I think that your teacher is trying to "baffle" you in an inappropriate manner. In my book, any number should be followed by the appropriate units. Conversely, if I ask someone to find a number, if the number is correct and the units are correct, then the answer is correct. I wish you fortitude, humor the guy and this too shall pass.

Or I could just kick his ***.