Math Check

[taibak]

Okay, I'm slightly embarassed to be asking this one where it's just a straight up, back of the envelope calculation based on Newton's Laws, but for the life of me, I can't get the numbers to work out.

I'm trying to find a rough estimate for how long it will take a spacecraft to reach the Moon based on the semi-major axis of the Moon's orbit and the top speed of Apollo 10.

I get 384,403,000m / 12,329 m/s = 31,179 s = 8.7 hours.

Granted, there's a lot of assumptions going into this, but I seem to be missing a couple of days here. Can't quite see how orbital mechanics would create that much of an error and numbers I'm starting with sound good. What am I missing here?

Thanx

[Bob B.]

12,329 m/s sounds a little high; I would have thought the number would have been closer to 11,000 m/s. Where did you get this figure?

The big problem is that this is the speed only at perigee. As soon as the spacecraft starts to move away from the Earth it slows down. It will continue to slow down until it reaches apogee and then it will begin to fall back toward Earth and speed up again.

The following Web page gives all the equations you need:

www.braeunig.us/space/orbmech.htm

Knowing the perigee distance and velocity you can calculate the apogee distance and eccentricity from equations 1.18 and 1.20. The semi-major axis is the average of the perigee and apogee distances. You can then calculate the true anomaly of the spacecraft as it crosses the Moon's orbit from equation 1.34. Finally, use the method shown in example problem 1.11 to calculate the the travel time.

[Bob B.]

I found some spare time to work out this problem. Using data from the web page,

history.nasa.gov/SP-4029/Apollo_18-24_Translunar_Injection.htm

we have the following at termination of the TLI burn,

Altitude = 1,093,217 ft (179.920 n mi)
Space-Fixed Velocity = 35,562.96 ft/sec
Flight Path Angle = 7.379 deg
Eccentricity = 0.97834

Let’s convert altitude to radius and switch to SI units. The equatorial radius of the Earth is 6,378.140 km, therefore the spacecraft’s distance from the center of the Earth at burnout is,

r = 6,378,140 + 1,093,217 / 3.28084 = 6,711,353 m

and the velocity is,

v = 35,562.96 / 3.28084 = 10,839.6 m/s

We can calculate the semi-major axis of the orbit using equation 1.36 from the web page,

www.braeunig.us/space/orbmech.htm

We have,

v = (GM x a (1 - e²))^0.5 / (r x cos phi)

where GM for Earth equals 3.986005E+14 m³/s², therefore

10,839.6 = (3.986005E+14 x a (1 - 0.97834²))^0.5 / (6,711,353 x cos(7.379))

a = 304,737,647 m

Using equation 1.34 we can calculate the true anomaly at TLI burnout and when the spacecraft crosses the orbit of the Moon. At burnout we have,

r = a x (1 - e²) / (1 + e x cos u)

6,711,353 = 304,737,647 x (1 - 0.97834²) / (1 + 0.97834 x cos u)

u = 14.84215 deg = 0.259044 radians

Apollo 10 entered lunar orbit on 21-May-1969 at 20:44:54 GMT. At this time the Moon’s distance from Earth was 404,351 km, therefore

404,351,000 = 304,737,647 x (1 - 0.97834²) / (1 + 0.97834 x cos u)

u = 171.54443 deg = 2.994015 radians

Using the method shown in example problem 1.11, we calculate the travel time as follows:

Equation (1.31),

cos E = (e + cos u) / (1 + e cos u)

Eo = arccos[(0.97834 + cos(0.259044)) / (1 + 0.97834 x cos(0.259044))]
Eo = 0.027256 radians

E = arccos[(0.97834 + cos(2.994015)) / (1 + 0.97834 x cos(2.994015))]
E = 1.911467 radians

Equation (1.33),

M = E - e x sin E

Mo = 0.027256 - 0.97834 x sin(0.027256)
Mo = 0.00059367 radians

M = 1.911467 - 0.97834 x sin(1.911467)
M = 0.989351 radians

Equation (1.30),

n = (GM / a³)^0.5
n = (3.986005E+14 / 304,767,647³)^0.5
n = 0.00000375246 rad/s

Equation (1.29),

M - Mo = n x (t - to)

t = to + (M - Mo) / n
t = 0 + (0.989351 - 0.00059367) / 0.00000375246
t = 263,496 sec = 3 days, 1 hr, 11 min, 36 s

According to the first link, TLI occurred at 19:28:20 GMT on 18-May-1969. Therefore the actual time from TLI to LOI was,

3 days, 1 hr, 16 min, 34 s

Drat, I’m off by 5 minutes!


Edit: Cleaned up equations to make more readable.

[taibak]

Makes sense, but I'm still stumped on how to find perigee distance.

Thanks for the help though.

-Taibak

[Bob B.]

taibak said:

Makes sense, but I'm still stumped on how to find perigee distance.

We know from NASA the eccentricity of the orbit was 0.97834, and I calculated the semi-major axis to be about 304,738 km. Therefore the perigee and apogee distances are,

Rp = 304,738 x (1 - 0.97834) = 6,601 km

Ra = 304,738 x (1 + 0.97834) = 602,875 km

[taibak]

Thanks again. I was trying to reinvent the wheel and calculate it from scratch. Totally confused.

-Taibak