Travel time for apollo to the moon with Kepler's thirs law

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Discussion Overview

The discussion revolves around calculating the travel time from Earth to the Moon using Kepler's third law. Participants explore the theoretical framework of orbital mechanics, the assumptions involved in the calculations, and the practical considerations of the Apollo missions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the travel time to the Moon as 4 days and 18 hours using Kepler's third law, based on the semi-major axis of the Earth-Moon distance.
  • Another participant points out that there are infinite possible elliptical trajectories that could affect travel time, questioning whether they all yield the same duration.
  • Concerns are raised about neglecting the effects of Earth's rotation, the relative positions of the spacecraft and the Moon, and the influence of the Moon's gravity on the trajectory.
  • A later reply suggests that the spacecraft's trajectory involves a combination of orbits, starting in Earth orbit and transitioning to Moon orbit, which complicates the application of Kepler's third law.

Areas of Agreement / Disagreement

Participants express differing views on the factors influencing travel time, with no consensus reached on the reasons for the discrepancy between the calculated and actual travel times of the Apollo missions.

Contextual Notes

Limitations include assumptions about the trajectory being a simple elliptical orbit, neglecting the complexities of gravitational interactions, and the initial conditions of the spacecraft's launch.

alex123go
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Hi everyone,

For one of my project in school, i need to calculate the travel time to the moon. To do that, I thought to use Kepler's third law which is :
(2(pi) / P )2 * a3 = GM

where :

P : period of travel
a : major semiaxes (not sure in english, but it is the half the longest radius of the ellipse) : in this case half of the earth-moon distance : 380 400km/2= 190 200km
G : gravitational constant : 6.67E-11

M : Earth's mass : 5,972E24 kg

Whit these values, my period is equal to 9 days and 13 hours, so the travel time to the moon is the half of that and is 4 days and 18 hours.

But in the apollo program, it takes about 3 days to go to the moon. Can someone help me to figure out why does I have almost 2 days of difference.

Is it because I didn't took notice of the moon gravity, because the spacecraft wasn't doing an ellipse like this or for another reason?

Thank you for your help
 
Engineering news on Phys.org
a : major semiaxes (not sure in english, but it is the half the longest radius of the ellipse)
In English: "semi major axis" ... the longest axis is the "major axis", and "semi-" means "half of" ... but "major semi-axis" also works.

There are infinite possible ellipses that would take a spacecraft from the Earth to the Moon - do they all have the same travel time?
You neglected the effect the rotation of the Earth would have and the relative initial positions of the craft and the moon ... as well as the Moon's gravity. What happens when the Moon's gravity is stronger than the Earth's? Does the scenario fit the requirements for Kepler's 3rd Law? You can find descriptions of the Apollo mission online to see how they did it.
 
Thank you for ou help. And I will look for the description on internet.
 
Consider that the spacecraft starts out in orbit about the Earth, and ends up in orbit about the Moon - traveling via the crossover point where Earth and Moon gravity cancel. Thus the trajectory is not going to be just one Kepler orbit - but a combination of them. They could get to the crossover point as quickly as they liked, and then burn fuel to sow down so the Moon can capture them.
 

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