# physics

If the average speed of an orbiting space shuttle is 19600 mi/h, determine the time required for it to circle the Earth. Make sure you consider the fact that the shuttle is orbiting about 200 mi above the Earth's surface, and assume that the Earth's radius is 3963 miles

1. 👍 0
2. 👎 0
3. 👁 1,175

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.

1. 👍 0
2. 👎 0
2. This was copied and pasted straight from my webassign problem set. I have spent 3 nights now trying to figure it out.

1. 👍 0
2. 👎 0
3. 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 = 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.

1. 👍 0
2. 👎 0
4. Sorry 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.

1. 👍 0
2. 👎 0
5. Considering 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.

1. 👍 0
2. 👎 0
6. The 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

1. 👍 0
2. 👎 0
7. A 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.

1. 👍 0
2. 👎 0

## Similar Questions

1. ### trig

A space shuttle 200 miles above the earth is orbiting the. Earth once every 6 hours. how far does the shuttle travel in one hour? Note the radius of earth about 4000 miles.

asked by Ali on September 17, 2011
2. ### Velocity and Displacement

The Space Shuttle travels at a speed of about 7.60 103 m/s. The blink of an astronaut's eye lasts about 120 ms. How many football fields (length = 91.4 m) does the Shuttle cover in the blink of an eye?

asked by Fainah on September 1, 2011
3. ### Physics

A 97kg astronaut and a 1100kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving it a speed of 0.13m/s directly away from the shuttle. Seven and a half seconds later the astronaut

asked by Arr-chan on May 27, 2014
4. ### Math

An observer 2.5 miles from the launch pad of a space shuttle measures the angle of elevation to the base of the shuttle to be 25 degrees soon after lift off. How high is the shuttle at that instant? (Assume that the shuttle is

asked by Jimmy on May 10, 2011
5. ### Physics

A 64.3 kg astronaut is on a space walk when the tether line to the shuttle breaks. The astronaut is able to throw a 10.0 kg oxygen tank in a direction away from the shuttle with a speed of 11.9 m/s, propelling the astronaut back

asked by Kev on January 26, 2015
1. ### Physics

the space shuttle is in a 250 mile high orbit. What are the shuttle's orbital period, in minutes, and it speed?

asked by Sarah on June 16, 2010
2. ### Physics

A(n) 57.3 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 54.3 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0.697 kg camera in her hand and

asked by Tina on January 22, 2015
3. ### Science

During a space shuttle landing, a parachute deploys from the back of the shuttle as it taxis down the runway. In addition to the speed brake, the parachute helps the shuttle come to a complete stop. How does the parachute help

asked by Connie on May 18, 2015
4. ### physics

A 88-kg astronaut and a 1200-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving it a speed of 0.25m/s directly away from the shuttle. Seven-and-a-half seconds later the astronaut

asked by ash on August 20, 2011