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

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Homework Help Overview

The problem involves calculating the speed required to maintain an orbit at the Earth's surface, considering the curvature of the Earth and the distance an object would fall. The context includes gravitational acceleration and the radius of the Earth.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss using the equation v=root(2gr) and question the validity of the resulting speed. They explore the time it takes for an object to fall a certain distance and how that relates to orbital speed.

Discussion Status

The discussion is ongoing, with participants questioning assumptions about the calculations and exploring different interpretations of the problem. Some guidance has been offered regarding the reasoning behind the calculations, but no consensus has been reached on the correct approach or answer.

Contextual Notes

Participants are considering the implications of Earth's curvature and the time it takes for an object to fall a specific distance, which may affect their calculations. There is a mention of imposed homework rules that may limit the depth of exploration.

Da Apprentice
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Homework Statement


At the 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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