Expert: Philip Stahl - 7/14/2008

Question
Hi, i am trying to do a review for a test on tuesday.... i cannot figure out the answers to 4 questions because i do not understand them and the concepts behind them......any help would be greatly appreciated......here they are

1)  What would be the angle between sun and southern horizon at noon on the winter solstice at 56o N latitude?

2)If mass of a satellite orbiting Earth would decrease by factor of 12 then gravitational pull from the Earth on that satellite would decrease by factor of ___.

3)If a satellite orbiting Earth would increase its altitude by factor of 9 then gravitational pull from the Earth on that satellite would decrease by factor of ____.

4)What speed would you need to achive a circular orbit around an asteroid of 1017kg of mass and 56 km in radius.

I am NOT looking for simply answers....answers will NOT HELP any because its merely a review to prepare for a test.  Answers would be appreciated for practice, but i really need an understanding of the formulas and how to use them....Thankyou in advance for any help you may be able to extend to me.

Answer
Hello,

My concern here is that first: you are sending this to me barely a day before your test (which puts more pressure on me to answer immediately) and also, what sort of learning curve is this? It is actually none! Zero! All the questions you raise are pretty basi stuff for any basic astronomy course, and your not knowing how to answer at this late stage is worrying in the extreme. Did you not go to the instructor at any point and ask for assistance? All the underlying concepts are fairly basic.

Let us take each in turn:

1) This question relies first on the basic geometry at work in the horizon system of coordinates. Also, knowing that the Sun's declination at the winter solstice is (-23.5) degrees. This means it must be 23.5 degrees south of the celestial equator.

So, what is the angle between the Sun at noon and the southern horizon if one is at latitude 56 N?

At noon, the Sun is on the meridian. Your meridian (I assume you know what this term means, otherwise please google it!) The latitude of 56 N means that at noon on your meridian, and directly overhead - will pass the declination of +56 degrees. The altitude of the Pole star (Polaris) from this location, will also be 56 degrees. Using a simple diagram (shaped like a semi-circle) you ought to be able to satisfy yourself from basic geometry that there will be 90 degress between the pole star position (on a horizon diagram) and the location of the celestial equator (at 0 degrees declination).

Note the celestial equator by this same geometry - is positioned 56 degrees south of the zenith (overhead point). Thus 0 degrees declination is now at (90 - 56) = 38 degrees to the southern horizon.

The Sun, meanwhile, as we saw - is located at (-23.5) degrees declination or 23. 5 degrees south of the CE. Then the angle between the Sun and southern horizon can easily be found from the geometry:

38 - 23.5 =   14.5 degrees

Sample question: repeat the problem but for a latitude of 13 degrees North (e.g. Barbados).

(2) is merely a simple application of Newton's law of universal gravitation, framed as the attractive force between earth (M_E) and the mass of the satellite (m):

F =  G (M_E)m/ r^2

where r is the distance of the satellite to the Earth's center.

Thus, the question asks how F will change if m becomes (m/12).

This is so elementary, given r, G, M_E are constants, it should not cause you two seconds of hesitation to solve. (Hint: you are using direct proportion: F ~ m)  If you can't solve this you shouldn't be taking this course.

(3) is slightly more difficult but similar principles apply.

The gravitational eqn. is simply modified to yield:

F =  G (M_E)m/ (R + h)^2

where R is the radius of Earth, and h is the altitude of the satellite. So, in this case, we are going from h to 9h in altitude. How will F change? Assume the numerator is constantt (k = G M_E m) and work on the basis of the proportionality and how it changes. This is basic math, not even astronomy!

4)  Again invokes the same principles but in a different scenario - now applied to an asteroid rather than Earth.

Here, we realize that the centripetal force of acceleration (of any satellite about the asteroid) is expressed in terms of the gravitational attraction of the satellite to the mass:

mv^2/ r  =  GM_A m/ r^2

where m is the satellite mass, M_A is the asteroid mass, and G is the gravitational constant (6.7 x 10^-11 N-m^2/ kg^2)and v is the velocity needed for orbit

Note the mass m cancels from both sides, so:

v^2/r =   G (M_A)/ r^2

or:

v = [G M_A/ r] ^1/2

then:

v = SQRT {(6.7 x 10^-11 N-m^2/ kg^2) (10^17 kg)/  5600 m)}

v = 34.6 m/s

NOTE: I changed the value of the asteroid mass to 10^17 kg since 1017 gave a preposterous (too low) answer. At least the one above makes sense.

Again, this is only a quick and dirty approach given the time limitations. What you *really* need is a full review session using a black board, complete with detailed diagrams and patient tutorial working. This (response) is better than nothing - but not much. In future try to get on top of these problems (principles) so you don't have to ask an expert one day (or two - but I only got it today)and keep your fingers crossed hoping for the best!