A minimum-energy transfer orbit to an outer planet consists of putting a spacecr

Question

A minimum-energy transfer orbit to an outer planet consists of putting a spacecraft on an elliptical trajectory with the departure planet corresponding to the perihelion of the ellipse, or the closest point to the Sun, and the arrival planet at the aphelion, or the farthest point from the Sun. (Assume the orbital radius of the Earth is 1.50 Times 1011 m, and the orbital radius of Mars is 2.28 Times 1011 m.) Use Kepler's third law to calculate how long it would take to go from Earth to Mars on such an orbit as shown in the figure above. Can such an orbit be undertaken at any time?

Explanation / Answer

A)The perihelion distance is the distance of the earth from the sun or 1 AU
the aphelion distance is the distance of Mars frm the sun or 1.52AU (this is an average value since Mars' orbit is more eccentric than the earth's)

therefore the major axis of the minimum energy orbit is 1+1.52=2.52 AU

the semi-major axis is then 1.26AU, and Kepler's third law tells us the period of an orbit with this semi-major axis is given by

P^2=a^3 where P is the period in yrs and a is the semi-major axis in AU, so we have

P^2=1.26^3 =>P=1.41 yrs

this is the time for the entire orbit; to get from the earth to mars would be half this value = 0.70 yrs

= 8.4 months

b) No, you could not just launch any time you wished; you would have to find the particular instance when Mars and the spacecraft would reach the desired aphelion rendezvous point at the same time

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