Question

MAVEN vs MOM
So i wanna talk about this in class and I wanna make sure is what i calculated and my concept does it really make sense or is it gibberish

Answer 1

First I ll start with MAVEN

it sat on a very powerful ATLAS V rocket
MAVEN also had a second stage CENTAUR which provides roughly (99 kN which i approximate to 100 kN of thrust)
mass of MAVEN (with fuel) = about 2500 kg
and i ESTIMATE that the mass of CENTAUR is roughly 6500 kg (Fully fuelled )..
its dry mass is about 2500kg too.. so i need a check on this u guys..

now MAVEN along with CENTAUR went into a low earth orbit .. which was circular and with a speed of 7.8km/sec

then the CENTAUR fired for 5 mins and shot it into heliocentric orbit all the way to mars. so here is where i do the calculation


\[I= F \Delta t = m \Delta v => 10^5 \times 300 =(2500 + 6500) \Delta v \]

\[\Delta v = \frac{3 \times 10^ 7}{9 \times 10^3} = 3.3 km/\sec\]

so.. the new velocity after burn time is 7.8 + 3.3 = 11.1km/sec.. close to what we need :D

so is this close enough..
cause i don't wanna use rocket equations and stuff

Answer 2

Next Comes MOM

Mass of about 1 ton (its a little more but i approximate it this way)
No CENTAUR.. it only has a baby rocket LAM .. so i don't consider its weight (again to keep things simple)

the thrust it could get is about 500 N. (extremely low).. and the rocket puts MOM with LAN in a very elliptical orbit to begin with

" India's first interplanetary spacecraft, was launched into an elliptical earth orbit with a perigee of 248.4 km and an apogee of 23,550 km"

http://www.isro.gov.in/pressrelease/scripts/pressreleasein.aspx?Nov07_2013

my first doubt over here is.. 248.4 km perigee is from the surface of the earth right? xD.. so when i calcluate i should add 6400 km to it right?

now.. i wanna know what speed this satellite will have at perigee? can some one tellm e that..


i want it to get somewhere close to 9.4 or 9. 6km/sec.. if its somewhere close to this.. then my rest of the calculations are correct

Plzz hlep me..

Answer 3

If .. that is true..

then the required delta v is 11.2 - 9.4 (lets say) 1.8 km/sec..

to achieve this.. we need a burn time of

\[\Delta t = \frac{m \Delta v}{F }= \frac{10^3 \times 1.8 \times 10^3 }{500 } = 3600\sec = 1 hr\]

that is exactly what is given .. a total burn time comes to 1 hr .. plz tell me these calcluations are sensible.. so i can discuss them..


and this 1 hr burn time is ridiculous to be done all at once.. so we do it stage wise in orbit raising manuever .. correct?

Answer 4

@ikram002p @Vincent-Lyon.Fr

Answer 5

@ganeshie8

Answer 6

all these names like MOM and MAVEN are distracting to the problem