Need help with transfer orbit time period

[HoboMoo]

I just don't even know where to begin. I'm not sure what formulas to use and just can't do anyhting with it. any help would be great. Thanks!

Recall that your trip to Mars is accomplished by using an elliptic transfer orbit going from Earth to Mars as shown in Fig. 1. This trajectory assumes that Earth at departure, the Sun, and Mars at arrival, are aligned. You calculated that the semi-major axis for this transfer orbit was a= 190100208000 m.

How long, in days, would the interplanetary trip last? Hint: first, determine the period of the transfer orbit.

 

[DrStupid]

How about Kepler's third law?

 

[HoboMoo]

So if a=190100208000m, its P^2=19010020800^3?

If that's the case, i get P^2= 6.8699 E 33 and square root that to get P?

P=8.2885 E 16?

 

[gneill]

Kepler's third law (in its original form) is a law of proportionality, not equality. To make it an equality you would have to use either a suitable (i.e. special) choice of units, a constant of proportionality, or form a ratio with another known pair of semi-major axis and period. So:

$$P^2 \propto T^3 $$
$$P^2 = k\;T^3$$
$$\frac{P2^2}{P1^2} = \frac{T2^3}{T1^3}$$
The last version is probably the easier to use if you happen to know of another suitable body orbiting the Sun for which you know the semi-major axis and the orbital period

 

[asteriatic]

To calculate the period of the transfer orbit, we can use Kepler's third law, which states that the square of the period (T) is proportional to the cube of the semi-major axis (a) of the orbit. This can be expressed as T^2 = (4π^2/GM)*a^3, where G is the gravitational constant and M is the combined mass of the Earth and Sun.

Plugging in the values for a and G (6.67x10^-11 m^3/kg/s^2), we can solve for T:

T^2 = (4π^2/6.67x10^-11 m^3/kg/s^2)*(190100208000 m)^3

T^2 = 5.182x10^19 s^2

T = 2.277x10^10 s

Converting this to days, we get approximately 662 days for the interplanetary trip to Mars. Of course, this is assuming a perfect alignment of the planets and a constant velocity throughout the trip. In reality, the trip may vary in duration depending on the launch window and the velocity of the spacecraft. I recommend consulting with a specialist or using more advanced simulations to get a more accurate estimate. Best of luck with your calculations!