Geosynchronous Puns

Do they have to burn fuel to do this? How much fuel and since they are so far away how far do they have to travel to stay in geosynchronous orbit.

Or do they have something built into the GPS protocols that accounts for the Earth's yearly tilt?

https://reddit.com/link/ru5ou8/video/8d265bkbm8981/player

Hello,

This is not homework help per se but I do need some help understanding something. I have been trying to work out geosynchronous orbits using the below format. My calculations came out basically correct and some googling determined that I had performed the calculations as I should have.

GMm/r^(2)=m(2πr)^(2)/rT^(2)

Sorry if the formatting is wonky, but I’m sure you know the equations I’m trying to express.

Anyway, my question is this. The mass of the satellite “m” on both sides cancels out, so essentially only the mass of the earth affects the radius of the orbit. How is this possible? What if the satellite were not of insignificant mass, something like a small black hole? Surely the mass of the satellite should affect the gravitational attraction and be included in the equation. Is this simply a case of classical mechanics being “close enough”, or am I missing something?

I hope someone can help me understand this. Thank you!

Are there any that remain in a fixed point above the earth's surface, or do they all wobble around?

I finally got enough gold after selling a crawling claw to craft a World Spinner rocket, which sold in 2 days and net me around 30k. On my server, they go for a bit more than 100kish each, but I've noticed that the Depleted Kpyarium Rocket sells for MUCH more than the World Spinner. While the world spinner only sales for 100kish, the depleted kpyarium sells for 160kish, even though both cost exactly the same to make. What am I missing here? Should I switch to Goblin engineering? Cause it kinda takes a fair bit of gold to switch over.

Maintaining geosynchronous orbit above target Citadel Enterprises standing by to deploy Low Orbital Ion Cannon for a third orbital bombardment.

First and second round fired and confirmed by ground forces. Massive damage to infrastructure and underground channels to outside sources. Target remains under seige.

Crew reports primary and secondary fuel cells are fully charged and green. Standing by for orders.

Does Mission Control confirm?

We all know that a geo(duna?)synchronous orbit is technically impossible because of ike, but is it possible to have a highly inclined and elliptical orbit to evade ike's orbit?

I mean obviously you could just use a heavy weight on the other end of the tether and move it in and out to compensate for different weight at the other end, but (assuming you want the half-g environment at the lower end) how far from the center of mass to the floor of the environment would the tether have to reach?

Or how about using a LaGrange point instead? Would the same trick work there?

Hey everyone.

I started playing WoW a few months ago and was always intrigued by the idea of paying for my account with gold. I've recently managed to make my first GWS because I found that engineering was a simple way to make big payouts.

My question is, if the US Median for an item like this is 200k or 300k, would posting it for 10-20k less make it much more likely to sell in your experience or is it roughly the same at price points like this? Basically should I worry about dropping my price for a quick sale or should I just wait, because it's still unlikely to sell quick?

Also, as an engineer is this my best way of making good gold or should I be looking at other options?

Thanks all and good luck out there!

P.S. Also, sorry if you're seeing this again, re-posting with flair.

A geostationary orbit is an orbit where the speed and distance above the Earth's surface allow the orbiting object to remain relatively "above" a single location on earth as it rotates.

How might I determine an objects distance from Earth's surface, knowing it's speed and that the orbit is geosynchronous?

If you can provide an answer, you can input your own speed. I'm not interested in a specific answer, only how to do such a calculation.

Here is some further information which i have found at brown.edu

Thank you!

“At an altitude of 124 miles (200 kilometers), the required orbital velocity is just over 17,000 mph (about 27,400 kph). To maintain an orbit that is 22,223 miles (35,786 km) above Earth, the satellite must orbit at a speed of about 7,000 mph (11,300 kph). That orbital speed and distance permits the satellite to make one revolution in 24 hours. Since Earth also rotates once in 24 hours, a satellite at 22,223 miles altitude stays in a fixed position relative to a point on Earth's surface. Because the satellite stays right over the same spot all the time, this kind of orbit is called "geostationary." Geostationary orbits are ideal for weather satellites and communications satellites.”

Boom boom time, boys. Let's get em.