Space elevator minimum initial speed

In summary: Should be:r_{GSO}=(\frac{GMT^2}{4*\pi^2})^{1/2}In summary, the conversation discusses the concept of a space elevator, which allows cheap access to space in the far future. Before this is possible, certain principles must be clarified. The first question (a) asks about the minimum initial speed needed for an object to escape the Earth's gravitational influence entirely, and this is solved using conservation of energy. The second question (b) asks about the height the elevator needs to have in order for an object released from it to ultimately escape Earth's gravitational influence. The third question (c) is similar to the first, but asks for the minimum initial speed needed for an object
  • #1
RiotRick
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0

Homework Statement


In the far future, humans have built a space elevator as a cheap
means of access to space. However before that could be done, a few basic principles had to be
worked out. . .

a)
What is the minimum initial speed (in an Earth-centered inertial reference frame) needed
for an object launching from the Earth surface (r = rE) to escape its gravitational influence
entirely? You can rely on conservation of energy.

b)
Objects can be launched into space from the elevator “just” by moving them up the elevator,
until a certain height, and then releasing them (with zero radial velocity). What is the
height ##r_{esc}## (as measured from the centre of the Earth) the elevator needs to have so that
the released object would ultimately escape the Earth’s gravitational influence? Careful:
the escape speed has not the same value as for the previous question.

Homework Equations


The previous question asked for the initial escape velocity.
##F_G = G*\frac{M*m}{r^2}##

The Attempt at a Solution



a) is solved via conservation of energy

b) Here I don't fully understand the question. The "careful" makes me suspicious. Do I set centrifugal force equal to the gravitational force or does it ask for the tangential velocity? So I'd have to set the tangential velocity equal to the escape velocity
##\omega*r=\sqrt(\frac{2GM}{r})##
 
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  • #2
First work out what the distance R from the Earth's centre must be for a free-falling geostationary satellite, by equating the formula for required centripetal acceleration (to maintain a circular orbit) to that for gravitational acceleration and solving for radius.

A free-falling object that momentarily has a geostationary velocity will escape the Earth if it is beyond the distance R, and will fall to Earth if it is less than that distance (Why?).

So if the top of the elevator is a satellite at distance beyond the distance R, tethered to Earth by the cable, the object can be released as soon as it is materially beyond distance R.
 
  • #3
I should have posted all the questions:

a)
Is there a circular orbit for which an object would hover continuously above the same
point on Earth? What is the radius rGSO of that geostationary orbit? Hint: for a circular
trajectory, the centrifugal force experienced by the object balances the gravitational pull
from the Earth.
b)
The space elevator starts at a base station on Earth at the equator, crosses geostationary
orbit and extends even further out into space. The whole structure revolves
with the Earth, at the same angular velocity. What is the direction of the force experienced
by an object positioned along the elevator below rGSO? Above?
c)
What is the minimum initial speed (in an Earth-centered inertial reference frame) needed
for an object launching from the Earth surface (r = rE) to escape its gravitational influence
entirely? Hints: in the worst case, the object’s speed would approach zero when r ! 1.
You can rely on conservation of energy.
d)
Objects can be launched into space from the elevator “just” by moving them up the elevator,
until a certain height, and then releasing them (with zero radial velocity). What is the
height resc (as measured from the centre of the Earth) the elevator needs to have so that
the released object would ultimately escape the Earth’s gravitational influence? Careful:
the escape speed has not the same value as for the previous question.

My solutions:
b) what you pointed out
a) ## r_{GSO}=(frac{GMT^2}{4*\pi^2})^{1/2) ##

c) ## v=\sqrt(\frac{2GM}{r_E}) ##

Now how is d) different from a) and b)?

I don't know what's wrong with my LaTex format
 
  • #4
The answers to (a) and (d) ARE radii, while the answer to (b) is a direction, so we only need to find a difference between (a) and (d). The answer is the same, but not tautologously so. An argument is needed to justify the conclusion that the greatest lower bound of radii at which a release with zero velocity relative to the elevator will lead to escape is the same as the geostationary radius.

RiotRick said:
I don't know what's wrong with my LaTex format
See the red parenthesis:

r_{GSO}=(frac{GMT^2}{4*\pi^2})^{1/2)

Also, the backslash is missing from before the frac.
 

Related to Space elevator minimum initial speed

1. What is a space elevator minimum initial speed?

A space elevator minimum initial speed is the speed required for a spacecraft or object to achieve a stable orbit when attached to a space elevator. It is the minimum speed needed to counteract the effects of gravity and maintain a consistent distance from the Earth's surface.

2. How is the space elevator minimum initial speed calculated?

The space elevator minimum initial speed is calculated using the equation v = sqrt(GM/R), where G is the gravitational constant, M is the mass of the Earth, and R is the distance from the center of the Earth to the desired orbit. This equation takes into account the gravitational pull of the Earth and the centripetal force needed to maintain a stable orbit.

3. What factors can affect the space elevator minimum initial speed?

The main factor that can affect the space elevator minimum initial speed is the altitude of the desired orbit. The higher the orbit, the higher the minimum initial speed required. Other factors that can affect the minimum initial speed include the mass of the spacecraft, the shape and length of the space elevator, and any external forces such as air resistance.

4. How does the space elevator minimum initial speed compare to traditional rocket launches?

The space elevator minimum initial speed is significantly lower than the initial speed needed for traditional rocket launches. This is because a space elevator takes advantage of the Earth's rotation and the centripetal force to achieve a stable orbit, whereas traditional rockets must use a large amount of fuel to overcome the Earth's gravity and achieve the necessary speed.

5. Can the space elevator minimum initial speed be decreased over time?

It is possible for the space elevator minimum initial speed to be decreased over time through advancements in technology. For example, the development of lighter and stronger materials could allow for longer and more efficient space elevator designs, reducing the minimum initial speed required for a stable orbit. However, it is unlikely that the minimum initial speed will ever be completely eliminated due to the laws of physics and the Earth's gravitational pull.

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