OT: Dumb Physics question

I was thinking about the orbital speed of a satellite around the earth which works out to 7900 meters per second according to my "Physics the Easy Way" book. They use the formula of V = square root (gr) where g is 9.8 meters per second and r is the radius of the earth or 6.4 million meters. But considering conservation of energy, the potential energy of a satellite with a mass of 1 Kg at 6.4 million meters height would be (mgh) or 62.72 million joules relative to the center of the earth. So, if it falls the distance of the radius of the earth, the kenetic energy will be 62.72E6 = 1/2 MV^2 and V will be 11,200 meters per second. Wiki has an interesting formula of v = square root of gm/r which makes no sense. So, why is the orbital speed around the earth 7.9 Km/s and not 11.2 Km/s ?

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Reply to
Bill Bowden
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g is smaller at that distance.

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Grizzly H.
Reply to
mixed nuts

"That distance" is the surface of the Earth.

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Rick
Reply to
rickman

I don't think the FAA would allow satellites to orbit at the surface of the earth.

v_orbital ~ v_escape/sqrt(2) when r = radius of the earth

If the kinetic energy >= potential energy, the orbit is open and v >= v_escape.

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Grizzly H.
Reply to
mixed nuts

** The PE of an object due to the earth's gravity can only be calculated with that simple approximation if it is both above and close to the earth's surface.

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.... Phil

Reply to
Phil Allison

g *inside* a uniform solid sphere is proportional to the distance from the center of the sphere. The kinetic energy works out to be the same.

--sp

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Best regards,  
Spehro Pefhany 
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Reply to
Spehro Pefhany

Thus, there's a speed (meters per second) value that can be formed from acceleration (g, measured in meters per second per second) and the orbital radius (meters). The units of g*r is the square of a velocity.

What does conservation of energy have to say about this? There's no energy relationship implied by an orbiting kilogram, and a long fall of a non-orbiting kilogram mass.

Orbital speed doesn't depend on the mass of the orbiting object; it depends on the mass of Earth, multiplied by G (which is not 'g' and is not meters per second squared). The value of 'g' is G*M/(r^2), where M is the mass of Earth, and 'r' is the distance from the center of the earth.

G is Newton's universal gravitational constant. It applies to every massive object. 'g' is local Earth surface gravity, given by a formula that includes the mass of Earth, and the distance of our usual abode from its center.

Reply to
whit3rd

ich

"

at what heigth?

ers

and here is the problem, r is not the radius of the heart, but the distance of the satellite from the earth center (radius of earth + heigth of satell ite deom earth surface)

ith

on

nope. It would be 0 (zero). "mgh" can be used only when near the earth surface, and h is the heigth _fr om_ earth surface. If the body is on the surface (h=0), his potential energy is 0.

When you calculate things in orbit you need to use Newton's universal gravi tation law where:

Epot = GMm/r (G gravitational constant, M mass of Earth, m mass of satell ite, r distance satellite center-earth center)

Bye Jack

Reply to
jack4747

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Must be something along the lines of the centripetal force mv^2/r being cou nterbalanced by the gravitational pull of mg, equate and solve for v=sqrt (gr). This is good enough for r not significantly larger than the earth rad ius. So looking at some numbers, say 200,000 ft altitude where air density supports such speeds without setting the house on fire, that is 200,000 / 3 .28/ 6350e3= 1% error, not enormously significant. btw sqrt(gr)=sqrt(9.8 x 6350e3)=7900 m/s is right.

Reply to
bloggs.fredbloggs.fred

The more precise formula is mv^2/r=GmM/r^2, G= grav. constant, M=earth mass, so v=sqrt(GM/r), this now makes sense if you correctly identify the variables.

Only Hawkins knows...

Reply to
bloggs.fredbloggs.fred

Whoops- should be Hawking...only Hawking knows.

Reply to
bloggs.fredbloggs.fred

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Cheers

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Syd
Reply to
Syd Rumpo

The FAA has no jurisdiction over satellites or anything else outside the US.

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Rick
Reply to
rickman

Oh. What if the satellites are really low (like at the level of tall grass) and their orbits take them over Kansas or Arizona or someplace like that and confuse air traffic control radars and the air traffic controllers too, and pose a potential terrorist threat?

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Grizzly H.
Reply to
mixed nuts

On Mon, 25 Jan 2016 15:25:17 -0500, mixed nuts Gave us:

17,000 mph.

That is the velocity of the ISS.

Reply to
DecadentLinuxUserNumeroUno

Sure it does. Everything above FL500 is Class A airspace.

Mars is under FAA (and related affiliates') jurisdiction. Fun trivia (I once trained for a pilot's license, long ago).

I expect they don't [actively] mind satellites, because once you've filed a flight plan (before launch), they tend not to change course. :-)

Tim

--
Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
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Reply to
Tim Williams

Many satellites can be moved around at will (at least until they run out of fuel).

Eg.

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Even some geosynchronous ones can be moved around.

--sp

--
Best regards,  
Spehro Pefhany 
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Reply to
Spehro Pefhany

Usually the potential energy is defined as zero at infinity, so the total energy for circular and elliptic orbits is always less than zero. For a parabolic "orbit" the total energy is always zero, while for a hyperbolic "orbit" it is greater than zero.

Reply to
upsidedown

On Mon, 25 Jan 2016 19:44:18 -0600, "Tim Williams" Gave us:

What about our specialized Laser Assassination Satellites!?

Reply to
DecadentLinuxUserNumeroUno

On Tue, 26 Jan 2016 02:15:44 -0500, Spehro Pefhany Gave us:

That is a very ambiguous remark.

They use fuel to attain and hold an orbital altitude to acquire a free fall condition. The do not "move around" much laterally as they get assigned to specific positions (geostationary), or orbits for those which traverse the globe. Those movements need to be known by all whom would use said 'local' space for their operations.

So, all of your "moving around" actions are done as corrections to retain prescribed behavior or positional stance.

Reply to
DecadentLinuxUserNumeroUno

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