Semimajor axis of orbit

The orbit of a spacecraft about the sun has a perihelion distance of 0.1
AU and an aphelion distance of 0.4 AU. I believe the semimajor axis of
orbit is 0.4 - 0.1 = .3 AU. Is this right? How do I get the orbital
period? I'm thinking of dropping this class because I don't see how to
solve the problems.


Thanks in advance

"scratch azazel" wrote...
in message ...

The orbit of a spacecraft about the sun has a perihelion distance of 0.1
AU and an aphelion distance of 0.4 AU. I believe the semimajor axis of
orbit is 0.4 - 0.1 = .3 AU. Is this right? How do I get the orbital
period? I'm thinking of dropping this class because I don't see how to
solve the problems.


Thanks in advance

'Lo Scratch --

I don't know, man, maybe you *should* give up...
this is pretty easy stuff...

On the other hand, the binary number system is a
snap as well, and it took me AGES to understand it.
Then one day a light turned on in my head and the
base-2 system became super-simple.

Okay, the semimajor axis of an ellipse can be found
usually in meters or astronomical units (AU's). And
it is equal to one-half of the major axis (the long axis)
of the ellipse. So if the aphelion distance from the
center of the Sun is 0.4 AU, and the perihelion
distance is 0.1 AU, then the major axis equals 0.5 AU
and the semimajor axis of the orbit is half that, or
0.25 AU.

As for the orbital period, you can find it using Kepler's
third law, again real easy math.

Watch those decimal points and units of measurement!
Don't let 'em get away from you.

Maybe if you keep at it, the photons will flash in your
mind, too?

happy days and...
starry starry nights!

--
Asimov! where have you gone?
Your written word goes on and on,
All becomes so clear to see
In Asimov's Astronomy!

Paine Ellsworth

"scratch azazel" wrote in message
...
The orbit of a spacecraft about the sun has a perihelion distance of 0.1
AU and an aphelion distance of 0.4 AU. I believe the semimajor axis of
orbit is 0.4 - 0.1 = .3 AU. Is this right? How do I get the orbital
period? I'm thinking of dropping this class because I don't see how to
solve the problems.

The major axis is the total distance of the line segment
from perihelion to aphelion. The semimajor axis is one
half that. Surely your textbook must have a diagram
showing an orbit and the axes?

After you have the semimajor axis, apply Kepler's law
for period.

"scratch azazel" wrote in message
...
The orbit of a spacecraft about the sun has a perihelion distance of 0.1
AU and an aphelion distance of 0.4 AU. I believe the semimajor axis of
orbit is 0.4 - 0.1 = .3 AU. Is this right? How do I get the orbital
period? I'm thinking of dropping this class because I don't see how to
solve the problems.

Finally! a question that even I can answer :-)

Looking thru some of the text books my kids used in University I found one
of Cosmology for nonscience majors, I am reading it now, I already went thru
Kepler and his laws, so here is my answer:

You really need to read your text book!! there is no way you will be able
to solve the problems if you don't know and understand the different terms
used and their real meanings (what is aphelion and perihelion in this case
for instance). So RT&@%$TB. Having said that, perihelion is the closest
distance from the orbit to the sun and aphelion is the farthest distance,
add them and you have the size of the major axis of the ellipsis describing
the orbit, half of that is know as the semimajor axis, which BTW, this
distance is equal to the average separation of the object in orbit to the
sun as it travel on its orbit around it (the sun). Once you know those
definitions, finding the semimajor axis is easy: add the perihelion plus the
aphelion and divide by two: (0.1 + 0.4)/2 = 0.25AU

For the next question, you'd need to memorize Kepler 3dr law: Kepler said
that the square of the orbital period is proportional to the cube of the
semimajor axis of the orbit, in other words: P^2 = K*R^3 where P is the
period, K is a constant and R is the semimajor axis. The same law, when
comparing 2 orbiting object is expressed as:

(P^2 / p^2) = (R^3 / r^3)

P and R and period and semimajor axis for one planet and p and r are for
another planet.
To resolve your problem, you need to know the semimajor axis of the
spacecraft (which we said it is 0.25AU), and you should also know the period
and semimajor axis of another planet, fortunately we know the one for Earth:
period = 1 year and semimajor axis = 1 AU , so the above formula becomes:

P^2 = R^3

and if you give the semimajor axis distance in AU units, the period will be
given in "Earth" years.

P^2 = 0.25^3

P = 0.125 years

Guillermo