PDA

View Full Version : astronomy HmK please help



science help me
2003-Oct-04, 09:22 PM
well i have this science project thing i must do and can all u mercury people help me please.


1.calculate how long it would take to travel from earth to the planet mercury at the speed of 36000 km/h, the fastest current technology

SarahMc
2003-Oct-04, 10:03 PM
Well, I'm not going to do your calculations for you, but I will give you a link to a site that gives you information on the planets.

The Nine Planets (http://www.seds.org/nineplanets/nineplanets/nineplanets.html)

And welcome to the forum! :D

science help me
2003-Oct-04, 10:40 PM
yes but what would i have to do to find the answer

ToSeek
2003-Oct-05, 01:34 AM
If your teacher would accept a simple answer, you can do the following:

- figure out how far out from the Sun Mercury orbits
- figure out how far out from the Sun Earth orbits
- get the difference
- divide by the speed

A more complicated (and realistic) answer would involve a Hohmann transfer orbit, but I don't know much about those except that it's the usual way to get from one planet to another.

science help me
2003-Oct-05, 02:51 AM
If your teacher would accept a simple answer, you can do the following:

- figure out how far out from the Sun Mercury orbits
- figure out how far out from the Sun Earth orbits
- get the difference
- divide by the speed

A more complicated (and realistic) answer would involve a Hohmann transfer orbit, but I don't know much about those except that it's the usual way to get from one planet to another.

so if mercury is 0.4 a.u
and earht is 1.0 a.u
would be 0.6 a.u
then it would be 21600

well is that right

Eroica
2003-Oct-05, 09:15 AM
so if mercury is 0.4 a.u
and earht is 1.0 a.u
would be 0.6 a.u
then it would be 21600


21600 what?

0.6 AU is about 90,000,000 km.
At a speed of 36,000 kph, it would take 2500 hours, or about 104 days, to cover this distance. [Distance = Speed x Time]

But this is simplistic. It assumes that the spaceship is starting from rest. But if the spaceship is launched from Earth, it's already travelling very fast along the Earth's orbit.

Eroica
2003-Oct-05, 09:20 AM
A more complicated (and realistic) answer would involve a Hohmann transfer orbit

True, but that would defeat the whole point of the exercise, since you can't arbitrarily choose to go at 36,000 kph if you're in gravitational freefall.

science help me
2003-Oct-05, 03:29 PM
so if mercury is 0.4 a.u
and earht is 1.0 a.u
would be 0.6 a.u
then it would be 21600


21600 what?

0.6 AU is about 90,000,000 km.
At a speed of 36,000 kph, it would take 2500 hours, or about 104 days, to cover this distance. [Distance = Speed x Time]

But this is simplistic. It assumes that the spaceship is starting from rest. But if the spaceship is launched from Earth, it's already travelling very fast along the Earth's orbit.

so the answer is 2500 hours, or about 104 days

Eroica
2003-Oct-05, 03:57 PM
I guess so.

R.A.F.
2003-Oct-05, 04:03 PM
so the answer is 2500 hours, or about 104 days

I'm not trying to sound like "that mean old guy" here, BUT...

While it's very important to arrive at the right answer, it's more important that you're able to understand how you determined that answer. In other words, giving you the answer doesn't really help you. I can see that others have given you the tools to help you...use them.

science help me
2003-Oct-05, 04:20 PM
yes but as u can see i did try it and i got it wrong so i thanx eroica so much

Ford Prefect
2003-Oct-05, 05:13 PM
I don't like to throw a spanner in the works. However even if you make the unrealistic assumption that the trip will be at a constant 36,000km/h is it reasonable to assume the distance is .6 AU, dont forget that during the trip Mercury will be continuing to travel in its orbit.

Eroica
2003-Oct-05, 05:25 PM
I was assuming that you time the trip to intercept Mercury at its closest approach. Of course, if it's on the other side of its orbit, you not only have much further to go, you also have to negotiate your way past the Sun.

Musashi
2003-Oct-05, 05:48 PM
Do you understand why you got it wrong? If you don't know how to get the right answer, just having the right answer isn't going to help you in the long run.

eburacum45
2003-Oct-05, 06:43 PM
104 days is strangely similar to the duration of a Hohmann transfer orbit, which is 105 days...

Matthew Ota
2003-Oct-05, 06:51 PM
At this Iowa State University web site:

http://www.polaris.iastate.edu/EveningStar/Unit6/unit6_sub2.htm

Hohmann transfer orbits have been used in the space program for years. It is the most fuel efficient trajectory to get to the planets. In effect, missions to Venus and Mercury involve "slowing down" the spacecraft so it falls in towards the sun. Earlier Mariner and Pioneer Venus probes used this method, and the future Mercury Messenger spacecraft will use it.

Quite an interesting study, to see how it is all done. Physics is fun!

Matthew Ota
Telescopes In Education 24 inch Cole Telescope Operator
http://tie.jpl.nasa.gov/tie/index.html

Eroica
2003-Oct-05, 08:26 PM
104 days is strangely similar to the duration of a Hohmann transfer orbit, which is 105 days...

Are you sure? I got 287 days.

Edit: Now I got 306 days! Help!

tuffel999
2003-Oct-05, 09:29 PM
Well hohmann transfer orbits aren't the fastest way between two planets just the most energy efficent(if you want ot call it that).

Basic answer to Hohmann vs. non-hohmann transfer based on engine type...
http://www-ssc.igpp.ucla.edu/dawn/mission.html

Technical explanation of Hohmann....
http://scienceworld.wolfram.com/physics/HohmannTransferOrbit.html

eburacum45
2003-Oct-05, 10:35 PM
Well, I did work it out using the proper equation, but as I don't trust my own maths I used this calculator as well
http://www.geocities.com/albmont/hohtransf.htm

which seems to give the same answer.

gethen
2003-Oct-05, 10:39 PM
Looks like "help me" got an answer he/she could use and signed off. Too bad. I don't think anyone here wants to pass out answers to homework questions. (I'm not faulting Eroica or anyone else for trying to help.) Any ideas about some way to deal with similar situations?

Eroica
2003-Oct-06, 08:54 AM
Well, I did work it out using the proper equation, but as I don't trust my own maths I used this calculator as well
http://www.geocities.com/albmont/hohtransf.htm

which seems to give the same answer.

Thanks for the useful link. You're right. I must have screwed up somewhere. But I don't understand how it can be so quick. Surely "motoring across the gap between the two orbits" would be quicker.

kucharek
2003-Oct-06, 09:08 AM
well i have this science project thing i must do and can all u mercury people help me please.


1.calculate how long it would take to travel from earth to the planet mercury at the speed of 36000 km/h, the fastest current technology

Welcome to the BABB, shm.

This is a poorly specified question. What does from-to means? What does 36000km/h means?

Is it surface-surface, surface-orbit, surface-flyby?
Is 36000km/h (10km/s) the delta-v capability of the whole propulsion system (launcher, upper-stage(s), spacecraft)?
Just for leaving Earth's gravity's sphere of influence needs roughly 11km/s delta-v. Then you've to brake away some of the 30km/s Earth orbital speed (you want to drop towards the sun, so you've to brake, seen relative to Earth). So what's these 36000km/h? You can only calculate a time when you give a trajectory you're on.

Harald

PS: It would be nice if you try to formulate your questions better and also show some work and thoughts you've already given into the subject. Otherwise it is easy to get the (wrong) impression that someone is just lazy and wants other people to do his work.

Eroica
2003-Oct-06, 09:13 AM
104 days is strangely similar to the duration of a Hohmann transfer orbit, which is 105 days...

\:D/ Eureka! I got out the lead and worked it out the oldfashioned way: 104.488 days.

That is a bit of a coincidence.

Jigsaw
2003-Oct-06, 01:05 PM
Any ideas about some way to deal with similar situations?
I hang out at another very large and busy science-stuff-cum-general-interest forum, and all such requests for homework help are always met with the blunt response, "We don't do people's homework for them. Next?"

If they're lucky, they'll get some kindhearted poster who will suggest ways to Google or otherwise get the information they need, but normally the response they get is fairly brisk.