analysis of clarke's geostationary orbit (geo), "surface" escape velocity, and the...

Preface

This article has been primarily based on my personal notes on the analysis of GEO orbit by considering the

gravitational force exerted on the satellite which provides the centripetal acceleration to maintain a circular

orbit around Earth.

At the time of writing, an online article1 mentioned the discovery of an “Earth-like” planet known as

COROT-7b located 150 parsecs away and that, “COROT-7b orbits its star [COROT-7] at a speed of more

than 750,000 km per hour, more than seven times faster than the Earth’s motion around the Sun”. This quote

from the article juggled around in my head and resulted in the culmination of an alternative analysis, which

proved the Earth’s orbital velocity about the Sun as approximately one-seventh of that of COROT-7b’s, and

lead to the derivation of a means for evaluating the orbital velocity without the need for stating an orbital

period.

Further musings on COROT-7b resulted in the discussion about escape velocity, which is significantly

related to many of the concepts discussed at the outset of this article; at least, the simple task of placing a

satellite in a geosynchronous orbit around the Earth.

— Michael de Silva, September 2009

Comments, Criticism, and Contribution

If you have any suggestions or discover errors in this document, write to [email protected] to

help make this document better. Thanks in advance for your contribution.

Contributors will be acknowledged in this document.

1http://www.independent.ie/world-news/scientists-find-first-earthlike-planet-outside-solar-system-1888840.html

Michael M. Wijetunge de SilvaMSc (Dist), BEng (Hons), MIEEE, MIEE, AMIMechE(UK), MASME

email: [email protected] • blog: bsodmike.com • resume: mwdesilva.com

1 Clark’s Geostationary Orbit (GEO)

This is an analysis utilising concepts from classical mechanics to establish the Geostationary Orbit (GEO).

The idea of a geostationary orbit was first published on a wide scale in a paper entitled “Extra-Terrestrial

Relays - Can Rocket Stations Give Worldwide Radio Coverage?” by Arthur C. Clarke, published in

Wireless World magazine in 1945. In this paper, Clarke described it as a useful orbit for communications

satellites. As a result this is commonly referred to as the Clarke Orbit (Hunley, Noordung, Stuhlinger and

Garland, 1995).

Figure 1: Illustration of the

satellite with mass m orbiting

the Earth of mass M at an an-

gular velocity of ω, a distance

r from the centre of the Earth.

It can be said that the centripetal acceleration required to maintain a circular

orbit is provided by the gravitational force, acting on the satellite. This can

be summed up as,

Fcentripetal = Fgravity (1)

From Newton’s second law of motion, this is equivalent to mac = mg

wherem is the mass of the satellite. As such, this reduces to the magnitudes

|ac| = |g|, and implies that the geostationary orbit is independent of the

mass of the satellite, and is determined by the point where the magnitude

of centripetal acceleration required for circular motion is equal to the

gravitational acceleration of free fall.

Following from the analysis of circular motion in classical mechanics, since

the angular velocity v = ωr, the magnitude of centripetal acceleration is

given as,

|acentripetal| =v2

r= ω2r (2)

In his law of universal gravitation, Newton states that two particles having masses m1 and m2 separated by

a distance r are attracted to each other with equal and opposite forces directed along the line joining the

1



particles. The common magnitude F of the two forces is,

F = G(m1m2

r2

)(3)

Considering the force the Earth exerts on an object, if the object has a mass m, and the Earth has mass M ,

and the object’s distance from the centre of the Earth is r, then the force that the Earth exerts on the object

is,

F = G

(mM

r2

)(4)

If we drop the object, the Earth’s gravity will cause it to accelerate toward the centre of the Earth. By

Newton’s second law (F = ma), the magnitude of this acceleration must equal |g| = (GmM/r2)/m, or,

|g| = GM

r2(5)

where M is the mass of the Earth, 5.9736 × 1024 kg, and G is the gravitational constant, 6.67428 ±0.00067× 10−11 m3 kg-1 s-2.

Therefore,

Fcentripetal = Fgravity

mω2r = GmM

r2

ω2r =GM

r2

r3 =GM

ω2→ r =

3

√GM

ω2= 3

√µ

ω2

(6)

2



where the standard gravitational parameter, µ = GM , and is known with much greater accuracy. It is

known as the geocentric gravitational constant for Earth, where µearth = 398, 600.4418± 0.0008 km3 s-2.

This formula could be generally applied to any planet if M = Mp, the mass of the planet in question, and

ω = 2π/Tp, where Tp would be the (sidereal) rotational period of the planet,

r =3

√GMpT

2p

4π2(7)

Considering the orbital period (the time taken for a single complete revolution around the Earth) to be one

sidereal day, or 86, 164.09054 s, the angular velocity ω is obtained (from ω = 2πf = 2π/T ) by dividing

the angle travelled in one revolution (360◦ = 2π radians) by the orbital period T,

r = 3

√µ

ω2=

3

√√√√√ 398.6004418× 103(2π

86.16409054× 103

)2 = 42, 164.1696 km(8)

Subtracting the Earth’s equatorial radius of 6, 378 km, gives an altitude of 35, 786.17 km for the Clark’s

orbit from the surface of the Earth. The total circumference of the orbit is 264, 925.29 km.

Evaluating the velocity of the satellite along this orbit,

v = ωr

=

(2π

86.16409054× 103

)· 42, 164.1696

≈ 3.0747.66 km/s ≈ 11, 068.776 km/h ≈ 6, 877.819 mph

(9)

3



1.1 An Alternative Evaluation of the Satellites Velocity Along Clark’s Orbit

An alternative derivation of the velocity of the satellite along this orbit may be obtained by a slightly

different consideration of Equation (1),

Fcentripetal = mvorbit

2

rorbit

= mgorbit = Fgravity (10)

From Equation (5)2 it can be seen that the gravitational force at an altitude of 35,786.96 km is only 2.285%

of that at the surface of the Earth,

gorbit =GMearth

rorbit2

=µearth

(6378 km · 6.611)2= 2.24199393× 10−1 m/s2 (11)

From Equation (10), an expression may be obtained for the orbital velocity of the satellite.

vorbit =√rorbit · gorbit =

√µearth

rorbit

= 3.0746 km/s (12)

However, it should be noted that the orbital period has not been considered up to this point in this alternative

analysis. From v = ωr two equations for the orbital period, T, may be obtained since ω = 2π/T ,

T =2πrorbit

vorbit

= 2π

√rorbit

3

µearth

= 86, 166.507 s(13)

The orbital period above is quite close to the value of a sidereal day, 86, 164.09054 s, but is not equal. This

is due to the lack of precision in the radius multiplier used for Clark’s Orbit, in the analysis above.

2The geocentric gravitational constant, µearth, has units km3 s-2. Hence, gorbit has been converted from km/s2 to m/s2 above.

4



Suppose a satellite could be put into orbit, right above the surface of the Earth, whilst ignoring air drag and

the terrific sonic boom that would accompany such an orbit. It would have a theoretical orbital period of

1.4 hours3 and an orbital velocity of vorbit = 7.91× 103 m/s.

If one were to consider the Earth’s orbital velocity about the Sun, the result could be obtained as,

v = ωr =2π

Tr =

2π

31.536× 106(1 AU) = 29, 805.66 m/s or 107, 300.36 km/h (14)

Similarly, Equation (12) may be used with the heliocentric gravitational constant, µsun. Hence, the orbital

velocity of the earth would be given by,

vorbit =

√(1 AU) · µsun

(1 AU)2= 29, 784.69 m/s (15)

Equation (15) provides a more accurate approximation as it does not consider the orbital period. The

orbital period in Equation (14), T , is taken as an approximation to a standard year of 365 days in the above

example.

2 “Surface” Escape Velocity of Earth

The term escape velocity can be considered a misnomer because it is actually a scalar (and not a vector), i.e.

it specifies how fast the object must move but the direction of movement is irrelevant, unless “downward”.

If a force acting on an object is a function of position only, it is said to be a conservative force, and it can

be represented by a potential energy function which for a one-dimensional case satisfies the derivative

condition,

−dUdx

= F (x) (16)

3http://hyperphysics.phy-astr.gsu.edu/HBASE/mechanics/earthole.html

5



Hence, the definition of potential energy, in terms of r,

U(r) = −∫ r

r0

F(r)dr (17)

with the choice of the zero of potential energy at infinite distance where the force approaches zero, the

gravitational potential energy is the work done to bring an object from infinity to radius r,

U = −∫ r

∞−mGM

r2dr = −mGM

r(18)

To derive the formula for escape velocity consider the use of conservation of energy. Imagine a space

shuttle of mass m is at a distance r from the centre of mass of the planet, whose mass is M . Its initial speed

is equal to its escape velocity, vescape. At its final state, it will be an infinite distance awgay from the planet,

and its speed will be negligibly small and assumed to be 0. Gravitational potential energy U and kinetic

energy K are the only types of energy concerned; as per the conservation of energy,

(U +K)i = (U +K)f

−mGM

r+

1

2mv2 = 0 + 0

(19)

It can be seen thatKf = 0 as the final velocity is assumed 0 and Uf = 0 as the final distance is approximated

to infinity. Hence, the escape velocity is,

vescape =

√2GM

r(20)

The orbit velocity for a circular orbit may be found by equating the gravitational force, Equation (5), to the

6

required centripetal force, Equation (2),

GM

r2=mv2

r→ vorbit =

√GM

r(21)

Note that the orbit velocity and the escape velocity from that radius are related by,

vescape =√

2 · vorbit (22)

On the surface of the Earth, the escape velocity is approximately 11,186.436 m/s or 40,271.17 km/h.

However, at 9,000 km altitude in “space”, it is slightly less than 7.1 km/s.

Escape velocity is independent of the mass of the escaping object. It does not matter if the mass is 1 kg

or 1,000 kg, escape velocity from the same point in the same gravitational field is always the same. What

differs is the amount of energy needed to accelerate the mass to achieve escape velocity: the energy needed

for an object of mass m to escape the Earth’s gravitational field is mGM/r, a function of the object’s mass.

More massive objects require more energy to reach escape velocity. All of this, of course, assumes we are

neglecting air resistance.

Trivia!

To evaluate the velocity at which the Earth’s surface rotates (on its own axes) where the Earth’s equatorial

radius (WGS84 ellipsoid) is taken as 6, 378.137 km,

v = ωr =2π

Tr =

2π

86, 400rearth = 463.82 m/s (23)

References

Hunley, J. D., Noordung, H., Stuhlinger, E. and Garland, J. (1995). The Problem of Space Travel: The

Rocket Motor, Diane Publishing Co.

analysis of clarke's geostationary orbit (geo), "surface" escape velocity, and the...

Documents