Find apogee distance, speed, and T of Earth satellite

[oddjobmj]

Homework Statement

An Earth satellite has a speed of 29300 km/hr when it is at its perigee, 246 km above Earth's surface. Find the apogee distance, the speed at apogee, and the period of revolution.

I thought I had this problem but I must be doing something silly because my answers are wrong.

Homework Equations

v=\sqrt{GM(2/r-1/a)}

from that:

a=\frac{GMr}{2GM-rv^2}

Radius at aphelion=r_a
Radius at perihelion=r_p

Semi-major axis=a=\frac{r_P+r_a}{2}

T²=\frac{4π^2r^3}{GM}

The Attempt at a Solution

I can find the semi-major axis from the following formula and my known values. I don't mind working symbolically but I will provide intermittent numerical results as checkpoints.

a=\frac{GMr}{2GM-rv^2}

a=7340 km

With the semi-major axis I can find the radius at aphelion=r_a:

a=\frac{r_P+r_a}{2}

r_a=2a-r_p

r_a=8066.5 km

With that I can find the velocity at aphelion=v_a:

v_a=\sqrt{GM(2/r_a-1/a)}

v_a=6673 m/s

I get the same thing if I simply use this relation:

\frac{v_a}{v_p}=\frac{r_p}{r_a}

I can also find the period using a=r:

T²=\frac{4π^2a^3}{GM}

T=6258 seconds

Where am I messing up? Thank you!

 

[Andrew Mason]

Try approaching this problem from an energy perspective. What can you say about the total energy? What does the total energy consist of? At what point in relation to the ellipse is the Earth located? What is the relationship between the perigee and apogee of an ellipse?

AM

 

[oddjobmj]

What can you say about the total energy? What does the total energy consist of?

The total energy is the kinetic energy of the satellite plus its potential energy due to the pull of gravity. The satellite's energy is mostly kinetic at the perigee and the alternative is true at the apogee.

At what point in relation to the ellipse is the Earth located?

The Earth is located at one of the foci.

What is the relationship between the perigee and apogee of an ellipse?

The distance between the perigee and apogee is greater than the distance between any two other points on the ellipse. Their average is the semi-major axis.

Perhaps the way I approached the problem is not the most elegant but is it wrong? I am more than happy to go about it a different way but the relationships I used should stand, no?

 

[Andrew Mason]

You haven't explained where you are getting your formula for v from so I don't understand your solution.

Since angular momentum (L = mvr) is conserved, you can determine the relationship between the r_a and r_b in terms of v_a and v_b. Is that where you are getting this from?

Using that, and the conservation of energy, you should be able to solve for r_a

AM

 

[D H]

You haven't explained where you are getting your formula for v from so I don't understand your solution.

It's a variation on the vis viva equation ($v^2 = GM(2/r - 1/a)$), Andrew, and it's a direct consequence of conservation of energy.


oddjobmj: What did you use for G*M and r? Did you do the units conversions properly?

 

[gneill]

Also: don't forget that the problem gives the height of the satellite above the Earth's surface at perigee, not its orbital radius.

 

[oddjobmj]

oddjobmj: What did you use for G*M and r? Did you do the units conversions properly?

I actually plugged the given units directly into wolfram alpha instead of doing any conversions. I have never had an issue with incorrect conversions and my resulting units came out correctly in each case. With wolfram alpha you can actually just type (mass of earth)*(gravitational constant) and it will plug in the values for you to a high degree of accuracy.

Also: don't forget that the problem gives the height of the satellite above the Earth's surface at perigee, not its orbital radius.

Yes, thank you. For r_p I used 246 km + radius of earth= 6613.5 kilometers

You haven't explained where you are getting your formula for v from so I don't understand your solution.

Since angular momentum (L = mvr) is conserved, you can determine the relationship between the ra and rb in terms of va and vb. Is that where you are getting this from?

Using that, and the conservation of energy, you should be able to solve for ra

AM

I am curious if you have tried this and found a different result than I did.

EDIT:

For example, my calculation for a:

http://www.wolframalpha.com/input/?...6613.5+kilometers)*(29300+kilometers/hour)^2)

 

[D H]

With wolfram alpha you can actually just type (mass of earth)*(gravitational constant) and it will plug in the values for you to a high degree of accuracy.

No, it doesn't. Your calculation used G*(earth mass). Wolfram Alpha is going to blindly accept what you wrote and use its own values. They are neither accurate nor precise nor consistent. G*(earth mass) is an observable quantity, accurate to 2 parts in a billion. in comparison, G and the mass of the Earth have an accuracy of about 1 part in 10 thousand -- if you are lucky and have a good source. Wolfram Alpha is not a good source in this regard.

You haven't been taught how utterly lousy the value of G is compared to other physical constants. For now, you should use the values in the back of your textbook, or in the chapter this homework problem comes from, or as specified in the home problem itself. The same goes for the mass of the Earth.

Yes, thank you. For r_p I used 246 km + radius of earth= 6613.5 kilometers

That means you used a value of 6367.5 km for the radius of the Earth. Where did you get that value from? Let me guess: Wolfram Alpha?

Use the value in the back of your textbook. It's most likely either ~6378.137 km (the Earth's equatorial radius) or ~6371.009 km (the mean radius, by a number of means). It almost certainly is not 6367.5 km.
Do not trust Wolfram Alpha or the google calculator to know physical constants if you are doing homework. Use the values from the back of your textbook. Like the values used in Wolfram Alpha or the google calculator, those values in that back of your text are almost certainly inaccurate, imprecise, and inconsistent with high precision measurements. They are, however, consistent with the answers you are expected to entry into your mindless, automated homework tool.

 

[oddjobmj]

No, it doesn't. Your calculation used G*(earth mass). Wolfram Alpha is going to blindly accept what you wrote and use its own values. They are neither accurate nor precise nor consistent. G*(earth mass) is an observable quantity, accurate to 2 parts in a billion. in comparison, G and the mass of the Earth have an accuracy of about 1 part in 10 thousand -- if you are lucky and have a good source. Wolfram Alpha is not a good source in this regard.

You haven't been taught how utterly lousy the value of G is compared to other physical constants. For now, you should use the values in the back of your textbook, or in the chapter this homework problem comes from, or as specified in the home problem itself. The same goes for the mass of the Earth.



That means you used a value of 6367.5 km for the radius of the Earth. Where did you get that value from? Let me guess: Wolfram Alpha?

Use the value in the back of your textbook. It's most likely either ~6378.137 km (the Earth's equatorial radius) or ~6371.009 km (the mean radius, by a number of means). It almost certainly is not 6367.5 km.



Do not trust Wolfram Alpha or the google calculator to know physical constants if you are doing homework. Use the values from the back of your textbook. Like the values used in Wolfram Alpha or the google calculator, those values in that back of your text are almost certainly inaccurate, imprecise, and inconsistent with high precision measurements. They are, however, consistent with the answers you are expected to entry into your mindless, automated homework tool.

Thank you for this. I will have to look them up because so far this semester (this is one of the last 3 homework problems of perhaps 100+) I haven't had an issue using the default alpha values. It is sort of silly that when I typed that they were accurate I knew someone would go after that claim. I just didn't realize how far off they might be.

After looking, I realize the book does not have values provided in an appendix. I might have to flip through related chapters to find values that were used in the book.

I will do some digging and try again with the new values. Does everything else look accurate?

 

[oddjobmj]

I found some values in various chapters of the book and used those in my equations and the results are still wrong.

M_E=5.976*10²⁴ kg
R_E=6370 km
G=6.67*10^-11

results:

r_a=8068 km

v_a=6674 m/s

period=6259 s

 

[Jediknight]

I actually plugged the given units directly into wolfram alpha instead of doing any conversions. I have never had an issue with incorrect conversions and my resulting units came out correctly in each case. With wolfram alpha you can actually just type (mass of earth)*(gravitational constant) and it will plug in the values for you to a high degree of accuracy.
Yes, thank you. For r_p I used 246 km + radius of earth= 6613.5 kilometers
I am curious if you have tried this and found a different result than I did.

EDIT:

For example, my calculation for a:

http://www.wolframalpha.com/input/?i=(gravitational+constant)*(mass+of+earth)*(6613.5+kilometers)/(2*(gravitational+constant)*(mass+of+earth)-(6613.5+kilometers)*(29300+kilometers/hour)^2)

thank you, I'm doing the same problem not sure if I got it my calculator just died but I feel so stupid for not thinking of the fact that there aint no satelite 121000m above Earth cm! its just me my textbook and my calculator (not wolfram savey and from the looks of it that's a good thing) but yes thatnk you very much!

 

[Jediknight]

Also: don't forget that the problem gives the height of the satellite above the Earth's surface at perigee, not its orbital radius.

just wanted to mention, this is what i wanted to quote and who I meant to thank, i guess I was tired and hit the wrong reply button yesterday

and with that help I solved problem today, ΔE=0 and ΔL=0 turned into (2*G*m(earth)/rp)-vp^2)ra^2-(2*G*m(earth)ra+(vp^2*rp^2) vp=velocity at perigee rp=radius at perigee from cm
im sure you could algebraically minipulate what the quadratic looks like, but solving a quadratic is the only way to solve this right?

w/ your numbers, (x=r(a))

http://www4b.wolframalpha.com/Calculate/MSP/MSP303721dfhe9b38e53b0600003ec247i27ica8cgb?MSPStoreType=image/gif&s=17&w=517.&h=85.

solving gives you x=6.624*10^6 m or 8.117*10^6 m, you know the former is your perigee distance so 8.12*10^6 m is your distance from Earth cm at apogee with that every answer should be fairly simple calculations

I hope I'm not disclosing too much here,(it seems like despite using wolfram alpha for source the O.P. is really trying to grasp the concept and at this point could use an answer to check against) please let me if I am

EDIT please let me know if I am not please let me if I am lol,also changed sig figs in answer since I didn't bother to look up Earth's mass to four

 

[gneill]

Yes, you're stuck with the quadratic unless you know some other handy relationships amongst the parameters of orbits. Then it becomes possible to use the given information to make your way through to the speed at aphelion and period of the orbit without solving a quadratic.

Unless you're course happens to be focused on astrodynamics it's unlikely that you would be expected to have memorized the formulas involved. So the "from scratch" approach using conservation of energy and angular momentum is the way to go.