A spaceship is fired into orbit from Cape Canaveral....

[beast]

I am having trouble with this problem. I am in High School Trig. The problem:

A spaceship is fired into orbit from Cape Canaveral. Ten minutes after it leaves the cape, it reaches its farthest distance north of the equator, 4000 km. Half a cycle later it reaches its farthest distance south of the equator (on the other side of the Earth) also 4,000 km. The spaceship completes and orbit in 90 minutes.

Ok so when I graph this I am having trouble with the sinusoidal . I think it would be the equator as its 4000 miles south and north of the cycle. But if it takes off from cape canaveral approx 12.5 minutes north of the equator is this correct?

If so where would my graph begin at the equator or?

If not where? I know the amplitudes must be the same so if i start at the point of cape canaveral how can I have equal amplitudes?

 

[stapel]

A spaceship is fired into orbit from Cape Canaveral. Ten minutes after it leaves the cape, it reaches its farthest distance north of the equator, 4000 km. Half a cycle later it reaches its farthest distance south of the equator (on the other side of the Earth) also 4,000 km. The spaceship completes and orbit in 90 minutes.

How long is a north-south-north cycle?

Note: An "orbit" refers to having made it all the way 'round the planet. But, as anybody who has watched the tracking of a satellite or orbitting craft knows, the path taken by the satellite is quite often not the same for each orbit. Each orbit is frequently shifted a bit from the last one, as can be seen by looking at the tracking for the International Space Station.

What is the actual question for this exercise? Are you supposed to find a trig model for the motion with relation to the equator, or something else?

The starting point of the graph will vary with the assumptions and the instructions. I rather doubt one would include the launch as part of the orbital path. But even were that to be included (with a lot of other assumptions), this would only cause a phase shift, wouldn't it?

Eliz.

 

[TchrWill]

I am having trouble with this problem. I am in High School Trig. The problem:

A spaceship is fired into orbit from Cape Canaveral. Ten minutes after it leaves the cape, it reaches its farthest distance north of the equator, 4000 km. Half a cycle later it reaches its farthest distance south of the equator (on the other side of the Earth) also 4,000 km. The spaceship completes and orbit in 90 minutes.

Ok so when I graph this I am having trouble with the sinusoidal . I think it would be the equator as its 4000 miles south and north of the cycle. But if it takes off from cape canaveral approx 12.5 minutes north of the equator is this correct?

If so where would my graph begin at the equator or?

If not where? I know the amplitudes must be the same so if i start at the point of cape canaveral how can I have equal amplitudes?

I assume that you are attempting to plot the height above the equatorial plane with time.

SInce you say 4000km in the body of the problem description and 4000 miles in a later statement, I am not certain as which you mean to use.

If a 90 minute orbit is correct, the altitude of the satellite must be 174 miles or 280km.

Assuming that 4000km (2486 miles)is what you intend:
The radius of the satellite from the center of the earth is 3963 + 174 = 4137 miles (mean earth radius = 3963 miles).

If the maximum height above the orbital plane is 2486 miles (4000km), the orbit inclination must be µ = arcsin(2486/4137) = 36.93º.

One quarter of the orbit is 90º or 22.5 minutes. Since the satellite reaches its maximum height above the equatorial plane 10 minutes after launch, liftoff takes place at 90(10/22.5) = 40º back from the maximum height point or 3963cos(40)tan(36.93) = 2282 miles above the equatorial plane.

If you want to plot height above the orbital plane vs time, at time = 0 and altitude = 0, the height, h, above the equatorial plane would be 2282 miles. Ten minutes later, h(max) = 2486 miles (4000 km). 22.5 minutes (90º) later, the satellite is on the equatorial plane. 22.5 minutes (90º) later, the satellite is 2486 miles (4000km) below the equatorial plane. 22.5 minutes later, it is back on the orbital plane and 90 minutes after reaching its maximum height, it is back at this point of 2486 miles (4000km).

The heights beyond the first maximum altitude point (t = 10 minutes, or µ = 40º), are defined by h = 4137cos(a)sin(36.93º, "a" being the orbital angle of the satellite from the maximum height point. For example, at a = 30º, h = 4137cos(30)sin(36.93) = 2152 miles. At a = 60º, h = 4137cos(60)sin(36.93) = 1243 miles. At a = 90º, h = 4137cos(90)tan(36.93) = 0. Without any further calculations, at a = 120º, h = -1243 miles. At a = 150º, h = -2152 miles. At a = 180º, h = -2486 miles.

I hope this is of some use to you.