Astronautics, Spring quarter 2003, HW 6, UCI

by Nasser M. Abbasi

1 problem 1, chapter 7. Weisle book

Problem: Given that total $ΔV = 4.29$ km/sec to transfer from LEO at inclination $280$ to GEO. $Isp = 453$ sec, $mp = 16,000$ kg, $ms = 1300$ kg. How much payload can tug deliver to GEO? can tug make a round trip without payload? if it can, how much payload could it carry to GEO and still return to LEO?

Assumptions

$m p$ given is that starting from LEO, and not from surface of earth.

roundtrip is back to LEO, not earth.

Method

We are given $ΔV$ and asked to find payload that could be carried given the physical properties of the spacecraft.

Solve from total $ΔV = V ln Z e$

For the other parts of the problem, use the rocket equation again to solve for different variables as shown in the analysis.

Analysis

$g = 9.8$ m/s.

so $V = I g = 453 (9.8) = 4439.4 e sp$ m/s

$ΔV = VelnZ$

hence $4290 = 4439.4 ln Z$

hence $lnZ = 44243990.4-= 0.9663468036$

hence $Z = e0.9663468036 = 2.628325112$

so $1+-λ= 2.628325112 ϵ+λ$

$λ$ is the payload ratio

$ϵ$ is the structural ratio

$λ = mmL+m-- s p$

$ϵ =--ms-- ms+mp$

so $--mL-- ms+mp+mL- 1+-λ= -m1s+ms+mpmL---= --mms++mmp-- = ms+mp+mL--= 1300+16000+mL- ϵ+λ ms+mp+ ms+mp- mss+mLp- ms+mL 1300+mL$

$13001+31060+00m0L+mL-= 2.628325112$

solve for $m L$

$1300+ 16000 + mL = 2.628325112$ $(1300)+ 2.628325112 mL$

$1.628325112 mL = 1300+ 16000- 2.628325112$ $(1300 ) = 13883.17735$

$mL = 11.38682833.1275173152-= 8526.047561 ≈ 8526$ kg

To find if tug can make a round-trip to LEO without payload, make $mp_new$ as the unknow, solve for it, and compare it to the given $m p$ .

Now we have an aditional $ΔV$ which is that needed to go back from GEO to LEO.

So, our $ΔV$ now is $4290+ 4290 = 8580$ km/s

$ΔV = VelnZ$

$8580$ $= 4439.4 ln Z$

hence $lnZ = 8580-= 1.932693607 4439.4$

hence $1.932693607 Z = e = 6.908092891$

$Z = mm0f-$

Here, $m0$ is the initial mass at start of the trip, which is $ms + mp_new$ , and $mf$ is the final mass at the end of the trip, which now is $m = m f s$

So, $m0- ms+mp_new- Z = mf = ms$ solve for $mp_new$ and compared to give $mp$ to see if less than.

$6.908092891 = 1300+m1p30_0new-$

$mp_new = 6.908092891(1300)- 1300 = 7680.520758kg ≈ 7680.5$ kg

Compare this to the $mp$ that the tug actually has which is $16,000$ kg, so the answer is Yes, it can make a round trip back to LEO with no payload.

To find how much payload it can carry and still make a round trip to LEO. Since the $mp$ needed to make a round trip with NO payload was found above to be $7680.5$ kg, then the $mp$ that we can use to make one second half of the round trip with no payload is $7680.5-= 3840.25 2$ kg

So, given that we started with $mp = 16000$ kg, then the $mp$ that we have at our disposal in the first half of the trip is the difference $16000 - 3840.25 = 12159.75$ kg. This is the $mp$ we can use for the one way trip from LEO to GEO with a payload. We know find this payload.

$ΔV = V lnZ e$

$4290 = 4439.4 ln Z$

hence $4290 lnZ = 4439.4-= 0.9663468036$

hence $Z = e0.9663468036 = 2.628325112$

$Z = ms+mL+mp--= 1300+mL+12159.75 ms+mL 1300+mL$ solve for $mL$

$1300+mL+12159.75 2.628325112 = 1300+mL$

$2.628325112 (1300)+ 2.628325112 mL = 1300 + mL + 12159.75$

$1.628325112 mL = 1300+ 12159.75 - 2.628325112 (1300)$

$m = 6167.642613 ≈ 6167.6 L$ kg

2 problem 7.4

see problem on page 226, Weisel book.

Assumptions: burn-time is zero long. g (earth accelaration) does not change during the flight of the spacecraft.

Method: Use the rocket equation

Analysis

for the overall system

$m0 = mp1 + ms1 + mp2 + ms2 + mL = 1167 + 113 + 415+ 41 + 250 = 1986$ kg

$mf = mL + ms2 = 250$ $+ 41 = 291$ kg

burn out for end of first stage:

$V = I g = 282 (9.8) = 2763.6 e sp$ m/sec

First stage:

$Vbo1 = VelnZ$

$m10 = mp1 + ms1 + mp2 + ms2 + mL = 1167 + 113+ 415 + 41 + 250 =$ $1986$ kg

$m1f = ms1 + mp2 + ms2 + mL = 41 + 415+ 41+ 250 = 747$ kg

$(m10) Vbo1 = 2763.6 ln m1f$

$( ) Vbo1 = 2763.6ln 1794867- = 2763.6ln2.6586 = 2702m/sec ≈ 2.7km/sec$

second stage:

$V = V lnZ bo2 e$

$m20 = mp2 + ms2 + mL = 415 + 41+ 250 = 706$ kg

$m2f = mL = 250$ kg

$( ) ( ) Vbo2 = 2763.6 ln mm20 = 2763.6 ln 720560 2f$

$Vbo2 = 2763.6ln 2.824 = 2869.043 ≈ 2869m/sec$

so, final burnout velosity is the sum of the above 2 velosities:

$2858.29 + 2869 = 5727.29m/sec$

To find max altitude with 250 kg.

Find the mechanical energy $E$ at surface of earth and at end of last stage, and use to solve for the unknowns $rmax$ since $E$ does not change over the path.

Convert $μearth$ to m/sec, which is

$( )3 ( )3 ( )3 3.986012 × 105 km/sec = ⇒ 3.986012 × 105 × 109 m/sec = 3.986012 × 1014 m/sec$

At surface of earth, and noting that the velosity of the rocket is zero at that point, we get $V2earth- --μ-- 0 -μ--- E = 2 - rearth = 2 - rearth$

$14 E = - reaμrth-= - 36.938768.011245××11003 = - 62494847$ m $2$ /s $2$

now, final velosity is $5727.29$ m/sec, so

$E =$ $V2max 2 -$ $-μ-- rmax$

$( 2 ) rμ--= Vma2x-- E = 57272.292- (- 62494847) = max$ $78895772$

$---μ--- 3.986012×1014- rmax = 78895772 = 78895772 = 5052250$ m

So, max altitude $5052250- 6378000 =$ $- 1325750$ m $= - 1,325.750$ km

Not sure why I get negative ALT. I think this is because zero potential energy reference is usually taken at $∞$ . This should then be

_________________________________max alt = $1,325.750$ km

At end of stage 1 we know the velocity. Let the spacecraft coast from that point untill its velosity becomes zero. Then start stage2.

At end of stage one, the mass of spacecraft is $m = 706 1f$ kg.

The K.E. the spacecraft have at this point is $2 0.5mV ,$ then at end of coast, this K.E. will all be exchanged by P.E. gained in going up, so solve for distance travelled

$12mV 2 = mgh$

$1(706)(2858.292) = 706 (9.8)h 2$

$h = 12(7067)(026(8958.8.)292)-= 416827m = 416.8km$

Now, the spacecraft fires its second stage rocket, at end of the second stage it will have gained a velosity of $2869$ m/sec (found from above). Mass of spacecraft at end of stage 2 is $mL = 250$ kg

From $12mV 2 = mgh$

$1V2 = gh 2$

$12V2- 0.5(2869)2 h = g = 9.8 = 419957m = 419.957km$