Launching a Rocket To mars help

[Digdug12]

Homework Statement

You are working on a project with NASA to launch a rocket to Mars, with the rocket blasting off from Earth when Earth and Mars are just aligned along a straight line from the sun. As a first step in doing the calculation, assume circular orbits for both planets. If Mars is now 60.1 degrees ahead of the Earth in its orbit around the sun, when should you launch the rocket?

Give your answer in days to the nearest whole number (i.e. 45.6 = 46)

Note: For this problem you need to know the fact that all the planets orbit the sun in the same direction, and the year on Mars is 1.72 Earth years.

Homework Equations

$$\theta$$(t)=.5$$\alpha$$t²+$$\omega$$_ot+$$\theta$$

The Attempt at a Solution

ok so first i used T=2pi/$$\omega$$ to find the $$\omega$$'s of Earth and Mars using 2pi/T=$$\omega$$,
$$\omega$$_e=1.99E-7
$$\omega$$_mars=3.43E-7
I also converted the initial position of Mars 60.1 degrees into radians, which is 1.049
I then set both position equations of $$\theta$$(t)=.5$$\alpha$$t²+$$\omega$$_ot+$$\theta$$ equal to each other and i got
1.049=t(3.43E-7 - 1.99E-7) and then proceeded to take that answer, divide by 60 for minutes, divide by 60 again for hours, and divide by 24 for the number of days.
The correct answer is 146, but i keept getting 84. What am i missing?

EDIT: Nevermind, i found my mistake, thanks anway!

 

[member38644]

It looks like you may have forgotten to include the initial position of Earth in your calculation. The equation for Earth's position should be \theta(t) = .5\alpha t^2 + \omega_e t + \theta_e, where \theta_e is the initial position of Earth. Since the problem states that Earth and Mars are aligned along a straight line from the sun, we can assume that their initial positions are the same. Therefore, your equation should be:

1.049 = t(3.43E-7 - 1.99E-7) + \theta_e

Solving for \theta_e and plugging it into the equation for Mars's position, we get:

\theta_mars(t) = .5(3.43E-7)t^2 + (3.43E-7)t + 1.049

Now, to find the time when the rocket should be launched, we set this equation equal to 0 and solve for t:

0 = .5(3.43E-7)t^2 + (3.43E-7)t + 1.049

Using the quadratic formula, we get t = 146.32 days. Rounding to the nearest whole number, the answer is 146 days. So it looks like you were on the right track, just missing the initial position of Earth in your calculation.