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2009-Feb-15, 08:27 AM
I hope someone can help with the calculations to this question in a textbook I am reading in anticipation of attempting a degree course in astronomy later this year.

The spacecraft is launched from the Earth and coasts along this ellipse until it reaches Venus when a rocket is fired either to put the spacecraft into orbit around Venus or to cause it to land on Venus.
A) Find the semi-major axis of the ellipse. Treat the orbits of both planets as circular.
B) Calculate how long in days such a one way trip to Venus would take."

Unless I'm missing something major, the semi major axis of such an ellipse should be 1 AU.

The answer given in the book is 144 days but no explanation of this is given. Any (simple) calculations would be appreciated.

Thanks,

Alan

agingjb
2009-Feb-15, 08:45 AM
Hints: I'd start by sketching the (three) orbits. Then, one of Kepler's laws.

tony873004
2009-Feb-15, 09:14 AM
Unless I'm missing something major, the semi major axis of such an ellipse should be 1 AU.

That's Earth's semi-major axis. Look up Venus' semi-major axis and average it with Earth's for your transfer orbit. Then use your sma to compute period.

Peter B
2009-Feb-15, 10:59 AM
I hope someone can help with the calculations to this question in a textbook I am reading in anticipation of attempting a degree course in astronomy later this year.

"The spacecraft is launched from the Earth and coasts along this ellipse until it reaches Venus when a rocket is fired either to put the spacecraft into orbit around Venus or to cause it to land on Venus.
A) Find the semi-major axis of the ellipse. Treat the orbits of both planets as circular."

Unless I'm missing something major, the semi major axis of such an ellipse should be 1 AU.

For what it's worth, that's what I think too. But I've never formally studied this stuff.

"B) Calculate how long in days such a one way trip to Venus would take."

The answer given in the book is 144 days but no explanation of this is given. Any (simple) calculations would be appreciated.

I think I can help you with this. What's the length in days of an Earth year? What's the length in days of a Venusian year? What's the average? See any relationship between that average and the answer given?

And welcome to the BAUT Forum.

2009-Feb-15, 11:44 AM
I think I've just found the answer.

To find the semi-major axis, you take the distance at perihelion i.e. the distance Venus is from the Sun which is 0.72 AU. Add to that the aphelion distance i.e the distance Earth is from the Sun which is 1 AU. The Earth Venus distance also has to be taken into account and it seems that half this distance is requried i.e. 0.14 AU. This gives a major axis of 1.86 AU with the semi-major axis being 0.93 AU or a bit over 139 million km.

The velocity a spacecraft requires to escape the Earth's gravity is a bit under 40,000 km and assuming that the spacecraft continues to move at this speed (Newton's 1st Law), this works out at around 146 days, the answer given in the book - not 144 days as I stated in my original post.

Alan

Fiery Phoenix
2009-Feb-15, 01:48 PM
Too lazy to calculate. I'm sure you got the right answer, though. Since it's the same number given in the book.

Jeff Root
2009-Feb-15, 05:17 PM

Does the question have a diagram? If so, does it show the elliptical orbit
of the spacecraft going from Earth on one side of the Sun to Venus on
the exact opposite side of the Sun? That would be a "Hohmann transfer
orbit", which is the most efficient way to go as far as fuel and energy are
concerned, but not necessarily the best choice in practice. Quicker trips
are possible, in which case the spacecraft makes less than half an ellipse
as it goes from Earth to Venus. The ESA spacecraft Venus Express took
153 days to get to Venus, covering a bit more than half an orbit, in order
to be in a better position to efficiently go into orbit around Venus.

-- Jeff, in Minneapolis

agingjb
2009-Feb-16, 08:47 AM
At the risk of being thought picky, dare I say that I suspect that Newton's First Law is being misapplied and that Kepler's Third Law would be more relevant in the OP's solution?

Bob B.
2009-Feb-20, 01:45 PM
Most of what you need to calculate orbits can be found here (http://www.braeunig.us/space/orbmech.htm). A quick and simple solution can be found using Kepler's third law, i.e. P^2~r^3.

Venus' distance from Sun = 0.72 AU
Earth's distance from Sun = 1.00 AU

Perihelion of Hohmann transfer orbit = 0.72 AU
Aphelion of Hohmann transfer orbit = 1.00 AU
Semi-major axis of transfer orbit = 0.86 AU

Period of Earth's orbit = 365 days

Period of orbit with semi-major axis of 0.86 AU:

P^2 / 365^2 = 0.86^3 / 1^3

P = SQRT[ 365^2 * 0.86^3 ] = 291 days

The transfer duration is 1/2 the orbital period, i.e. 291/2 = 145.5 days.

In reality, true Hohmann transfers don't work because the orbits of the Earth and the target planet are not in the same plane. It is, therefore, typical to intersect the target planet a little before or a little after the true half-way point. For instance, a Hohmann transfer to Mars should take about 8.5 months, but most spacecraft flying to Mars generally have durations of about 7 months or 10 months (give or take a little). If the intersection takes place before the half-way point, it is called a Type I transfer, and if after the half-way point, a Type II transfer.

Here is a diagram illustrating a Type I transfer:

http://www.braeunig.us/pics/fig9B2.jpg

Another option is to launch initially into the plane of Earth's orbit. After the spacecraft has traveled approximately 1/4 of the way around the Sun, a plane change is performed to incline the orbit so that it intersects Mars.