Time to Mars (orbit around Sun given perihelion and aphelion)

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SUMMARY

The discussion focuses on calculating the time for a satellite's journey from Earth to Mars, specifically in an elliptical orbit with a perihelion of 1 AU and an aphelion of 1.52 AU. The semimajor axis of the orbit is determined to be 1.26 AU. The participant initially struggles with calculating travel time due to the changing radius of the orbit and the non-linear nature of gravitational acceleration. Ultimately, they reference Kepler's Third Law, confirming that the ratio T²/R³ for the satellite matches that of Earth, leading to a calculated travel time of approximately 516.83 days, which is then adjusted to account for the journey to Mars.

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Homework Statement



Flight to Mars. To send a satellite from Earth to Mars, a rocket must accelerate the satellite until it is in the correct elliptical orbit around the sun. The satellite does not travel to Mars under rocket power, because that would require more fuel than it could carry. It just moves on a Keplerian orbit under the influence of the sun's gravity. The satellite orbit must have perihelion r = 1 AU, and aphelion r = 1.52 AU.

What is the semimajor axis of the satellite's orbit? (this works out to 1.26 AU; answer is correct)

Calculate the time for the satellite's journey. Express the result in days.

Homework Equations


ω(t)=2∏/T
v(t)=R*dω/dt-R2

The Attempt at a Solution



My brain goes a little wonky when I try to calculate travel time without a velocity.

I tried using G*M/r^2=a and plugging in the mass of the sun and the gravitational constant as well as our semi-major axis as r to find the acceleration due to gravity of the sun. Of course the radius is changing and due to the square the acceleration is not a linear function so using the semi-major axis (effectively an average) is highly inaccurate. I'm just not sure what else to use. You could differentiate the function but I don't know how to represent the radius as a function (an ellipse) given only the perihelion and aphelion.

Once you find the acceleration due to gravity you can plug it into our a=R*(dω/dt)-R*ω^2 and solve for T (ω=2*pi/T). Using what I have, though, the answers I have come up with are not correct. (516.3 days)

Does anyone have any suggestions?
 
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What does Kepler's third law tell you?
 
Thank you!

It tells me that the T^2/R^3 of the satellite should be the same as for the Earth which is 2.977E-19. It worked out to 516.83 which is actually really close to my original answer but I then had to divide that by 2 to get the time until arrival.
 

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