What is the Speed Needed for Orbit at Earth's Surface?

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To achieve orbit at Earth's surface, one must travel at approximately 8000 m/s, as the Earth curves downward at a rate that would drop away 5 meters over an 8 km horizontal distance. Initial calculations suggested a speed of 11,179.4 m/s, which was deemed excessively high. The discussion highlighted the importance of understanding the relationship between the time it takes for an object to fall and the required speed for maintaining orbit. Participants emphasized the need for a "sanity check" on calculations to ensure they align with physical principles. Ultimately, the consensus is that a speed of around 8000 m/s is necessary for orbital motion at the surface level.
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Homework Statement


At teh Earth's surface, the curvature is such that if you traveled in a straight line at a tangent to the surface of the Earth for 8kms the Earth would drop away 5 m. How fast would you need to travel to be in orbit at the Earth's surface?

(radious of the Earth 6.37*10^6m)

Homework Equations


I was thinking of using v=root(2gr).


The Attempt at a Solution


v= root(2*9.81*6.37*10^6)
= 11,179.4m/s

(I figured this was way to fast)
 
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Da Apprentice said:

Homework Equations


I was thinking of using v=root(2gr).
Can you explain your reasoning? How long does it take for an object to fall 5 m?

AM
 
If you drop an object initially at rest and 5 meters above the earth, how much time does it take the object to fall 5 meters?
 
SammyS said:
If you drop an object initially at rest and 5 meters above the earth, how much time does it take the object to fall 5 meters?

the Earth curves so that is you move at a tangent to the Earth the Earth curves downwards 5metres ... you would be above the Earth by 5metres. If this is so how fast would you need to travel to be in orbit at the Earth's surface?


AM said:
Can you explain your reasoning? How long does it take for an object to fall 5 m?

it would take about 1/2 a second, what's your point?
 
Last edited:
Da Apprentice said:
... = 11,179.4m/s

(I figured this was way to fast)
How did you figure?
it would take about 1/2 a second, what's your point?
Your estimate is not close. AM's point, if I may assert, is to help you think about your problem in a way that will not only provide a solution but also a "sanity check" of your method/result that you somehow think is questionable.
 
Da Apprentice said:
it would take about 1/2 a second, what's your point?
No. Work it out using s = at^2/2

So if it takes off horizontally 1 metre above the Earth surface and travels 8 km in exactly the time it takes to drop 5 m, how far above the Earth will it always be (assume the Earth is a perfect sphere)?

AM
 
Ok Thanks so you'd need to be traveling at about 8000m/s.
 
Yup .
 

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