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August 30, 2016
Posted by **savy** on Wednesday, September 7, 2011 at 11:10pm.

- physics -
**tchrwill**, Thursday, September 8, 2011 at 10:56amUsing your numbers:

T = 2(3963+200)5280(3.14)/19,600 = ?

However, the velocity of a satellite orbiting at 200 miles altitude is only 17,256mph.

If your intent was that the orbit is elliptical, you would need the perigee and apogee velocities to average 19,600mph. Since the perigee velocity is already only 17,256mph, and the apogee velocity is less, an average of 19,600 is impossible.

Might you have provided an incorrect number. - physics -
**savy**, Thursday, September 8, 2011 at 11:44pmThis was copied and pasted straight from my webassign problem set. I have spent 3 nights now trying to figure it out.

- physics -
**tchrwill**, Friday, September 9, 2011 at 11:07amThe velocity required to keep a satellite in a "circular" orbit derives from

Vc = sqrt(µ/r) where Vc = velocity in feet per second, µ = the earth's gravitational constant = 1.407974x10^16 ft.^3/sec.^2 and r = the orbital radius in feet.

Therefore, the circular velocity required to keep a satellite in a 200 mile high orbit (r= (3963+200)5280 = 21,980,640 feet is Vc = 25,309 fps = 9rp17,256 mph.

A velocity of 19,600 mph implies an orbital radius of r = µ/V^2 = 1.407974x10^16/28,746^2 = 17,038,831 feet = 3227 miles, inside the earth's radius. NOT POSSIBLE.

If the average velocity of 19,600 mph is meant to be the average of the perigee and apogee velocities of an elliptical orbit, you would have to derive an elliptical orbit having a perigee velocity higher than 17,256 mph and an apogee velocity that averages 19,600mph when added to the perigee velocity.

The perigee velocity is defined by

Vp = [2µ(Ra/(Ra+Rp)]^1/2 while the apogee velocity is defined by

Va = [2µ(Rp/(Rp+Ra)]^1/2.

You will have to play with these expressions until you find an elliptical orbit whose perigee and apogee velocities average 19,600 mph. - physics -
**tchrwill**, Friday, September 9, 2011 at 12:46pmSorry for the typo.

The velocity required to keep a satellite in a "circular" orbit derives from

Vc = sqrt(µ/r) where Vc = velocity in feet per second, µ = the earth's gravitational constant = 1.407974x10^16 ft.^3/sec.^2 and r = the orbital radius in feet.

Therefore, the circular velocity required to keep a satellite in a 200 mile high orbit (r= (3963+200)5280 = 21,980,640 feet is Vc = 25,309 fps = 17,256 mph.

A velocity of 19,600 mph implies an orbital radius of r = µ/V^2 = 1.407974x10^16/28,746^2 = 17,038,831 feet = 3227 miles, inside the earth's radius. NOT POSSIBLE.

If the average velocity of 19,600 mph is meant to be the average of the perigee and apogee velocities of an elliptical orbit, you would have to derive an elliptical orbit having a perigee velocity higher than 17,256 mph and an apogee velocity that averages 19,600mph when added to the perigee velocity.

The perigee velocity is defined by

Vp = [2µ(Ra/(Ra+Rp)]^1/2 while the apogee velocity is defined by

Va = [2µ(Rp/(Rp+Ra)]^1/2 where Ra and Rp = the apogee and perigee radii respectively.

You will have to play with these expressions until you find an elliptical orbit whose perigee and apogee velocities average 19,600 mph.

The perigee velocity must be lower than 24,403 mph or the satellite will escape earth's gravity, never to return. - physics -
**tchrwill**, Friday, September 9, 2011 at 12:58pmConsidering that the perigee velocity must be less than 24,403 mph, if it were 24,402 and the companion apogee velocity very small, say 1 mph, the average of the two would be far less than 19,600 mph, indicating that it is impossible.

- physics -
**Savy**, Friday, September 9, 2011 at 1:05pmThe radius of the Orbit about the earths center is 200 + 3963 = 4163miles.

Circumference of the orbit is 2ðR = 2*ð*4163 = 26156.9 miles

The average speed of an orbiting space shuttle is 19 800 mi/h,

Hence it takes a time of 26156.9miles / 19 800 mi/h, = 1.32 hours

Figured it out - physics -
**tchrwill**, Friday, September 9, 2011 at 2:14pmA typical Space Shuttle mission orbit is ~185 miles. The orbital velocity required to maintain this orbit is 25,355 fps = 17,287 mph, not 19,800 mph.

If, for whatever reason, your problem source chooses to make use of a totally ficticous orbital velocity, they should state so.

The "real" time for a spacecraft to orbit the earth at a circular altitude of 185 miles is 90.45min.

The "real" time to orbit at 200 miles altitude is 90.94min.