# physics

posted by .

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

• physics -

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 -

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

• physics -

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.

• physics -

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.

• physics -

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.

• physics -

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

• physics -

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.

## Similar Questions

1. ### physics

The average speed of an orbiting space shuttle is 20100 mi/h. The shuttle is orbiting about 196 mi above the Earth’s surface. Assume the Earth’s radius is 3963 mi. How long does it take to circle the Earth?
2. ### 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?
3. ### physics

The Earth’s radius is about 6380 km. The space shuttle is orbiting about 300.1 km above Earth’s surface. If the average speed of the space shuttle is 25700 km/h, ﬁnd the time required for it to circle Earth. Answer in units …
4. ### Physics

The De-orbit Burn The Shuttle must reduce its velocity at a pre-calculated point in its orbit in order to return to Earth. In order to reduce the velocity and change the orbit of the Shuttle, a maneuver called the de-orbit burn is …
5. ### Math

Radius of earth 3,959 miles Space shuttle orbits 200 miles above earths Surface Space shuttle completes 220 degrees of a Revolution around earth in one day What is the average speed of the space Shuttle
6. ### Math

Radius of earth 3,959 miles Space shuttle orbits 200 miles above earths Surface Space shuttle completes 220 degrees of a Revolution around earth in one day What is the average speed of the space Shuttle
7. ### physics

The Earth’s radius is about 6380 km. The space shuttle is orbiting about 275.0 km above Earth’s surface. If the average speed of the space shuttle is 23500 km/h, find the time required for it to circle Earth.
8. ### Physics

The shuttle must perform the de-orbit burn to change its orbit so that the perigee, the point in the orbit closest to Earth, is inside of Earth's atmosphere. De-orbit maneuvers are done to lower the perigee of the orbit to 60 miles …
9. ### Physics

The shuttle must perform the de-orbit burn to change its orbit so that the perigee, the point in the orbit closest to Earth, is inside of Earth's atmosphere. De-orbit maneuvers are done to lower the perigee of the orbit to 60 miles …
10. ### Physics

The average speed of an orbiting space shuttle is 19700 mi/h. The shuttle is orbiting about 125 mi above the Earth’s surface. Assume the Earth’s radius is 3963 mi. How long does it take to circle the earth?

More Similar Questions