weightlessness and satellites in orbit orbital energies€¦ · 4/11/2016  · orbital energies...

Orbital energies Weightlessness and satellites in orbit

Review

• PE = - 𝐺𝑀𝑚

𝑅

• 𝑣𝑒𝑠𝑐𝑎𝑝𝑒 = 2𝐺𝑀𝐸

𝑅 = 2𝑔𝑅𝐸

• Keppler’s law: 𝑅3

𝑇2 = 𝐺𝑀𝑠

4𝜋2

Orbital Motion • Orbital velocity ≠ 𝑒𝑠𝑐𝑎𝑝𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

• In orbital motion an object is still under the influence of Earth’s gravitational field, which causes it to follow a parabolic path

• 𝐹𝑐= 𝐹𝑔

𝑚𝑣2

𝑅𝐸 =

𝐺𝑀𝐸𝑚

𝑅𝐸2

𝑣2= 𝐺𝑀𝐸𝑚𝑅𝐸

𝑅𝐸2

v = 𝐺𝑀

𝑅𝐸 = 𝑔 𝑅𝐸

If the elevator is accelerating upward at 2 ms-2, what does Dobson observe the dropped ball’s acceleration to be? SOLUTION: Since the elevator is accelerating upward at 2 ms-2 to meet the ball, and the ball is accelerating downward at 10 ms-2, Dobson observes an acceleration of 12 ms-2. If the elevator is accelerating ↓ at 2, he observes a = 8 ms-2.

Consider Dobson inside an elevator which is not moving…

If he drops a ball, it will accelerate downward at 10 ms-2 as expected.

What is weightlessness?

The ball is NOT weightless, obviously. It is merely accelerating at the same rate as Dobson! This is what we mean by weightlessness in an orbiting spacecraft

If the elevator is accelerating downward at 10 ms-2, what does Dobson observe the dropped ball’s acceleration to be? SOLUTION: He observes the acceleration of the ball to be zero! He thinks that the ball is “weightless!”

Considder an elevator moving down at 10𝒎𝒔−𝟏

Weightlessness

• Astronaut and spaceship have the same acceleration

• Acceleration is towards the center of the planet

• Both fall at the same rate

• There is no reaction force between astronauts and the spaceship

We have all seen astronauts experiencing “weightlessness.” Explain why it only appears that they are weightless. SOLUTION: The astronaut, the spacecraft, and the tomatoes, are all accelerating at ac = g. They all fall together and appear to be weightless.

International Space Station

Question

Only in deep space – which is defined to be far, far away from all masses – will a mass be truly weightless.

In deep space, the r in FG = GMm

𝑅2 is so large

for every m that FG, the force of gravity, is for all intents and purposes, zero.

Satellites in Orbit Geostationary Orbit

• Above plane of equator

• Does not move when viewed from Earth

• Receiving dish on Earth does not need to track transmitting antenna on satellite

• Communication uses

Polar Orbits

• Close to Earth’s surface (100km above it)

• Orbit is over the poles

• Over 24 hours satellite will view every point on Earth’s surface

Geosynchronous satellite

• Orbital period is 24 h

• Stays in the same area in the sky but wanders a figure 8 shape

Example • A geostationary satellite has an orbital time of 24h.

A) Calculate the distance of the satellite from the surface of the Earth B) Find the gravitational field strength at orbital radius

Example • A satellite wishes to

orbit Earth at a height of 100km above the surface of the earth. Find the speed and acceleration of the satellite.

• Solution

v = 𝐺𝑀

𝑅

• v =6.67𝑥10−11𝑥5.98𝑥1024

6.38𝑥106+100𝑥103

• = 7.85 x 103m𝑠−1

• a = 𝐺𝑀

𝑅2 or a = 𝑣2

𝑅

• a = 9.53m𝑠−2

Satellite energy

• Two speeds: 𝑣𝑒𝑠𝑐𝑎𝑝𝑒 and 𝑣𝑜𝑟𝑏𝑖𝑡

~ 11.2km𝑠−2 ~ 7.8 km𝑠−2

• Orbital shape depends on speed: 𝑣𝑜𝑟𝑏𝑖𝑡 at 7.8 circular 7.8 < v < 11.2 elliptical 𝑣𝑒𝑠𝑐𝑎𝑝𝑒at 11.2 parabola v > 𝑣𝑒𝑠𝑐𝑎𝑝𝑒 hyperbola

An orbiting satellite has both kinetic energy and potential energy.

The gravitational potential energy of an object of mass m in the

gravitational field of Earth is PE = – GMm

𝑅 , where M is the mass of the

earth.

The kinetic energy of an object of mass m moving at speed v is

KE =1

2mv2.

Thus the total mechanical energy of an orbiting satellite of mass m is

total energy of an orbiting satellite

Energies of satellites

𝑇𝐸 = 𝐾𝐸 + 𝑃𝐸

𝑇𝐸 =1

2mv2 -

GMm𝑅

Show that KE of an orbiting satellite having mass m at a distance R from

the center of Earth (mass M) is KE = GMm

2𝑅

SOLUTION:

In circular orbit 𝐹𝑐 = Fg

𝑚𝑣2

𝑅 =

GMm𝑅2

mv2 = GMm

𝑅

1

2mv2 =

GMm2𝑅

kinetic energy of an

orbiting satellite KE = GMm

2𝑅=

1

2

GMm𝑅

= 1

2 𝑃𝐸

Example

Three expressions for KE

• Since GM = 𝑔0 𝑅𝐸2

• KE = GMm

2𝑅=

𝑔0 𝑅𝐸2𝑚

2𝑅

In summary

• KE = 1

2 PE =

GMm2𝑅

= 𝑔0 𝑅𝐸

2𝑚

2𝑅

The IBO expects you to derive these relationships.

Show that the total energy of an orbiting satellite at a distance r from

the center of Earth is TE = – GMm

2𝑅

SOLUTION: From TE = KE + PE and the expressions for KE and PE we have

TE = KE + PE

TE = GMm

2𝑅 –

GMm𝑅

TE = GMm − 2GMm

2𝑅

TE = – GMm

2𝑅

total energy of an orbiting satellite PE= –

GMm𝑅

, KE = GMm

2𝑅, 𝑇𝐸 = −

GMm2𝑅

Example

Graph the kinetic energy vs. the radius of orbit for a satellite of mass m about a planet of mass M and radius R.

SOLUTION: Use KE = GMm

2𝑅. Note that KE decreases with increasing radius.

It has a maximum value of KE = GMm

2𝑅

EK

r R 2R 3R 4R 5R

GMm 2R

Graphical representation of KE

total energy of an orbiting satellite

Graph the PE vs. the radius of orbit for a satellite of mass m about a planet of mass M and radius R.

SOLUTION: Use PE = –GMm

𝑅. Note that PE increases with increasing

radius. It becomes less negative.

EP

r R 2R 3R 4R 5R

GMm R

-

total energy of an orbiting satellite

Graphical representation of PE

FYI Kinetic energy (thus v) DECREASES with radius.

Graph the total energy TE vs. the radius of orbit and include both KE and PE.

SOLUTION:

GMm R

-

E r R 2R 3R 4R 5R

GMm 2R

-

GMm 2R

+

EK

EP

Graph of TE

Thus a spacecraft must SLOW DOWN in order to reach a higher orbit!

total energy of an orbiting satellite

TE of a satellite is always negative Remember: Friction causes TE to decrease

• R decreases

• As R ↓, 𝑣 ↑

• As v ↑, 𝐾𝐸 ↑

• When would TE be zero?

• Far from gravitational pull so that PE = 0 and if motionless, KE = 0, hence TE = 0

• If KE = PE the object has just enough energy to move to distance where PE = 0, but no KE would be left motionless

Conclusion • If TE = 0

can just escape from central object

• If TE > 0 can escape and keep going

• If TE < 0 cannot escape and is bound to planet. This is called binding energy BE

• The extra energy needed to free the object is the BE.

• BE = - TE

= - ( - GMm

2𝑅 )

= GMm

2𝑅

Summary

• 𝐾𝐸 = 𝑇𝐸 = 𝐵𝐸

• 𝑃𝐸 = 2 𝐾𝐸 = 2 𝑇𝐸 = 2 𝐵𝐸

• For a satellite close to a planet 𝑅𝑜𝑟𝑏𝑖𝑡 = 𝑅𝑝𝑙𝑎𝑛𝑒𝑡 = 𝑅

• KE = GMm

2𝑅

• On planets surface BE to break free is BE = GMm

2𝑅

total energy of an orbiting satellite

Example • Calculate the minimum energy required to put a

satellite of mass 500kg into an orbit that is at a height equal to the Earth’s radius above the surface of the Earth.

• Solution

• ∆𝑉𝐸= 𝑔0𝑅𝐸 and ∆𝑉𝑎𝑡 𝑅 =𝑔0𝑅𝐸

2

𝑅

• ∆ V = 𝑔0𝑅𝐸 - 𝑔0𝑅𝐸

2

𝑅 = 𝑔0𝑅𝐸 1 −

𝑅𝐸

𝑅

• R = 2 𝑅𝐸

• ∆ V = 𝑔0𝑅𝐸 1 −𝑅𝐸

2 𝑅𝐸 = 𝑔0𝑅𝐸 1 −

1

2 = 𝑔0𝑅𝐸

2

• W = ∆ Vm = 𝑔0𝑅𝐸𝑚

2=

10𝑥6.4𝑥106𝑥 500

2= 16000MJ

• In orbit it will have KE, hence TE = PE + KE = 16000 x 106+ 𝑔0𝑅𝐸

2𝑚

2𝑅 = 16000 x 106+

𝑔0𝑅𝐸2𝑚

4𝑅𝐸 = 16000 x 106 +

10𝑥6.4𝑥106𝑥 500

4= 24000MJ

KE is POSITIVE and decreasing.

GPE is NEGATIVE and increasing (becoming less negative).

Example

From Kepler’s 3rd law, T 2 = [ 42/ (GM) ] r3.

Example

Thus r3 = [ GM / (42) ]T 2.

That is to say, r3 T 2.

From Kepler’s 3rd law T 2 = [ 42 / GM ]r3. Then

T = { [ 42/GM ]r3 }1/2

T = [ 42 / GM ]1/2r 3/2

T r 3/2.

Example

From Kepler’s 3rd law TX2 = (42 / GM)rX

3.

From Kepler’s 3rd law TY2 = (42/ GM)rY

3.

TX = 8TY TX2 = 64TY

2.

TX2 / TY

2 = (42 / GM)rX3 / [(42 / GM)rY

3]

64TY2 / TY

2 = rX3 / rY

3

64 = (rX / rY)3

rX / rY = 641/3 = 4

Example

Since the satellite is in uniform circular motion at a radius r and a speed v, it must be undergoing a centripetal acceleration.

Example

Since gravitational field strength g is the acceleration, g = v2/ r.

R

x

F = mg = GMm / x2 = mv2/ x.

Thus v2 = GM / x. Finally v = GM / x.

Example

R

x

But EK = (1/2)mv2.

Thus EK = (1/2)mv2 = (1/2)m(GM / x) = GMm / (2x).

EP = mV and V = –GM / x.

Then EP = m(–GM / x) = –GMm / x.

From (a), v2 = GM / x.

…continued

R

x E = EK + EP

E = GMm / (2x) + –GMm / x [ from (b)(i) ]

E = 1GMm / (2x) + -2GMm / (2x)

E = –GMm / (2x).

…continued

R

x

The satellite will begin to lose some of its mechanical energy in the form of heat.

…continued

R

x

Refer to E = –GMm / (2x) [ from (b)(ii) ].

If E , then x (to make E more negative).

If r the atmosphere gets thicker and more resistive.

Clearly the orbit will continue to decay (shrink).

…continued

total energy of an orbiting satellite

E = – GMm / (2r) EK = GMm / (2r) EP = – GMm / r

If r decreases EK gets bigger.

If r decreases EP gets more negative (smaller).

Topic 10: Fields - AHL 10.2 – Fields at work

Orbital motion, orbital speed and orbital energy

Escape speed is the minimum speed needed to travel from the surface of a planet to infinity.

It has the formula vesc2 = 2GM / R.

Example

To escape we need vesc2 = 2GM / Re.

The kinetic energy alone must then be E = (1/2)mvesc2 =

(1/2)m(2GM / Re) = GMm / Re. This is to say, to escape E = 4GMm / (4Re).

Since we only have E = 3GMm / (4Re) the probe will not make it into deep space.

…continued

Recall that EP = –GMm / r.

Thus ∆EP = –GMm ( 1 / R – 1 / Re ).

…continued

The probe is in circular motion so Fc = mv2/ R.

But FG = GMm / R2 = Fc.

Thus mv2/ R = GMm / R2 or mv2 = GMm / R.

Finally EK = (1/2)mv2 = GMm / (2R).

…continued

The energy given to the probe is stored in potential and kinetic energy. Thus

∆EK + ∆EP = E

GMm / (2R) – GMm(1/ R – 1/ Re) = 3GMm / (4Re) 1 / (2R) – 1 / R + 1 / Re = 3 / (4Re)

1 / (4Re) = 1 / (2R) R = 2Re.

…continued

It is the work done per unit mass by the gravitational field in bringing a small mass from infinity to that point.

Example

COMPARE: The work done by the gravitational field in bringing a small mass from infinity to that point is called the gravitational potential energy.

The phrase only differs by omission of “per unit mass”.

V = –GM / r so that V0 = – GM / R0.

But –g0R0 = –(GM / R02)R0 = – GM / R0 = V0. Thus V0 = – g0R0.

…continued

0.5107 = 5.0106 = R0.

At R0 = 0.5107 clearly V0 = -4.0107.

From previous problem g0 = –V0 / R0

= – -4.0107 / 0.5107

= 8.0 m s-2.

Vg = (-0.8 - -4.0)107 = 3.2107

∆EK = – EP

EK – EK0 = – EP

0

(1/2)mv2 = ∆EP

v2 = 2 ∆EP / m

v2 = 2 ∆Vg v2 = 2(3.2107)

v = 8000 ms-1.

This solution assumes probe is not in orbit but merely reaches altitude (and returns).

…continued