SPH4UI Gravitational Potential energy Mr. Burns. To explore how much energy a spacecraft needs to escape from Earth’s gravity, we must expand on the topic

SPH4UI

Gravitational Potential energyMr. Burns

To explore how much energy a spacecraft needs to escape from Earth’s gravity, we must expand on the topic of gravitational potential energy.

To calculate the change in gravitational potential energy that a mass undergoes when it undergoes a vertical displacement near the Earth’s surface, we used: GU mg h Where is the change in gravitational potential energy, m is the mass, g is the magnitude of the gravitational field constant, and is the vertical displacement, where when we choose hi to be the ground, we simply obtain U=mgh. This equation assumes that g remains reasonably constant during the change in vertical displacement.

GUh

Gravitational Potential Energy

We define the work done on a system by a specific conservative force as the negative of the change in a potential energy function associated with the force. Changes are always taken as final minus initial values

conserv f iW U U

U


As we recall from the Kinetic energy Theorem:

change in the Kinetic net work done on

energy of a particle the particle

K Kf KiE E E W A force does positive work when it has a vector component in the same direction as the displacement. If the work done on a particle is positive, then the particles kinetic energy increases.

Since the work done a we move a mass between two points is independent upon the path, the force is conservative

conserv KW U E

U W

f

iW F x dx

f

iU F x dx

f

i

h

f ihU mg dh mgh mgh

Change in potential energy is equal to the negative of the work done by the gravitational force on a mass

Calculus definition of work

This gives us a Calculus definition of change in potential energy

Equation for change in potential energy

U mgh If we set, hi to be zero (the ground), we obtain an simple formula to determine the potential energy


Work done by the gravitational force FG during the vertical motion of a mass from initial height y1 to final height y2 as the body rises. The gravitational potential energy U increases (negative work done by gravity, Kinetic Energy decreases).

Work done by the gravitational force FG during the vertical motion of a mass from initial height y1 to final height y2 as the body rises. The gravitational potential energy U decreases (positive work done by gravity, Kinetic Energy increases)


But, how do we handle Gravitational Potential Energy, when g is not a constant (such as when we are not on Earth)?


Recall that the law of universal gravitation is given by:2G

GMmF

r

To increase the separation of the two masses requires work to be done. That is, we must do work to overcome the gravitational attraction between the two masses when moving mass m from distance r1 to distance r2

Where is the magnitude of the force of gravitational attraction between any two objects, M is the mass of one object, m is the mass of the other object, and r is the distance between the centres of the two

GF

M mr1


r2

Work is done to overcome this distance

As the result of the work being done to increase the separation from r1 to r2, the gravitational potential energy of the system increases.

The work done to change the separation from r1 to r2 is equal is equal to the change in gravitational energy from r1 to r2. However, recall that work done by a varying force is equal to the area under the force-displacement graph for that interval.


To find an expression for this work, we consider a body of mass m outside the Earth and compare the work Wgrav done by the gravitational force when a body moves directly away from or towards the centre of the Earth (from r=r1 to r=r2).

The path does not have to be a straight line, it could also be a curve, but the work done only depends upon the initial and final values of r, not the path taken (this is definition of a conservative force).


2

1

2

212 1

r rE E E

grav r r

Gm m Gm m Gm mW Fdr dr

r r r

We now define the corresponding potential energy U so that Wgrav=U1-U2. Therefore we have a final definition of gravitational potential energy

1 2GmmU

r

Because the Force points directly

toward the centre of the Earth, F is

negative

You may be troubled by the equation for gravitational potential energy because it states that it is always negative. But we have encountered this before, recall when we used U=mgh, we found that U could be negative whenever the body of mass was below the arbitrary height we chose h to be zero at.


1 2GmmU

r

In using our equation, we have chosen U to be zero when the body of mass is infinitely far away from the Earth .

As the body moves towards the Earth, gravitational potential energy decreases and becomes negative.

r

These formula seen initially confusing when first encountered.

U mgh 1 2GmmU

r

When gravity field is constant

(near Earth)

Universal Potential energy, Equals zero when r approaches infinity

But, they provide us with the same meaning. The change in gravitational energy at the Earth's Surface is just a special case of the general solution. This is the gravity well scientists talk about.


What is the change in gravitational potential energy of a 72.1 kg astronaut, lifted from Earth’s surface into a circular orbit at an altitude of 4.35 x 102 km?

2 211 24 11 24

2 2

6 5 6

9 9

6.67 10 5.98 10 72.1 6.67 10 5.98 10 72.1

6.38 10 4.35 10 6.38 10

4.2198 10 4.5076 10

2

f i

f i

U U U

GMm GMm

r r

N m N mkg kg kg kg

kg kg

m m m

J J

8.88 10 J

66.38 10Er m 245.98 10EM kg


5

8

72.1 9.80 4.35 10

3.07 10

U mgh

Nkg m

kg

J

Let’s approximate the answer using U=mgh

Note: h is distance above

the ground


We can use the fact that if no other external forces besides gravity affect a mass, then the total energy is conserved (the gravitational force does work and thus the mechanical energy is conserved). The Kinetic Energy gained by the system is matched with the Potential Energy lost, and thus our mass will have the following statement of energy conservation..

1 1 2 2K KE U E U Now by inserting the formulas for the Kinetic and Potential Energy and interpreting the scenario where the object reaches the r location where both EK2 and U2 have a zero value (where the object comes to a stop).

2

2

1

22

2

GMmmv

rGM

vr

GMv

r

1 1 2 2

1 1

2

0

10

2

K K

K

i

E U E U

E U

GMmmv

r

This gives us the initial speed v1, needed for a

body to escape from the surface of a spherical mass M with radius r

This is the escape velocity of the mass


Suppose you want to place a 1000 kg weather satellite into a circular orbit 300 km above Earth’s surface.a) What speed, period, and radial acceleration must it have?b) How much work has to be done to place the satellite in orbit?c) How much additional work must be done to make this satellite escape Earth’s gravity

RE=6380 kmME=5.97x1024 kg

First we need the radius of the satellites orbit:

66380 300 6680 6.68 10r km km km m



For a stable orbit: FG=mac

2

2E

E

GM m vm

r r

GMv

r

2

11 242

6

6.67 10 5.97 10

6.68 10

7720

N mkg

kgv

mm

s

66.68 10r m



For a stable orbit: FG=mac

2

2 2

2 2

6

4

4

2

2 6.68 10

7720

5440

90.6min

E

E

GM m rm

r T

r rT

GM

r

v

m

ms

s

66.68 10

7720

r m

mv

s



2

2

6

2

7720

6.68 10

8.92

c

va

r

msm

m

s

66.68 10

7720

r m

mv

s


b) How much work has to be done to place the satellite in orbit?The work required is the difference between the total mechanical energy when the satellite is in orbit (Ef) and when the satellite was on the launch pad (Ei).

2

2 211 24 11 24

2 2 2

6 6

10

2

6.67 10 5.97 10 1000 6.67 10 5.97 10 10001

1000 77202 6.68 10 6.38 10

f i

Kf f Ki i

E E

E

W E E

E U E U

GM m GM mmv

r r

N m N mkg kg kg kg

m kg kgkg

s m m

10 10

10

2.99 10 6.25 10

3.26 10

J J

J

2

2

1

2

1

2

circular orbit2

f Kf f

E

E E

E

E E U

GM mmv

r

GM GM mm

r r

GM m

r

RE=6380 kmR =6680 km

ME=5.97x1024 kg


c) How much additional work must be done to make this satellite escape Earth’s gravity

• We recall that for a satellite to escape to infinity, the total mechanical energy must be zero.

• The total mechanical energy in orbit Ef was -2.99 x 1010 J• To increase this to zero, an amount of work equal to 2.99 x 1010 J must be

done.


A 5.00x102 kg weather satellite is to be placed into a circular geosynchronous orbit (one orbit is 24 hours) around Earth .

a) What is the radius of the satellite’s orbit?b) What is the velocity of the satellites orbit?c) What is the gravitational potential energy (universal) of the satellite when it is at rest on Earth’s surface?d) What is the total energy of the satellite when it is in geosynchronous orbit?e) How much work the launch rocket do on the satellite to place it into orbit?f) Once in orbit, how much additional energy would the satellite require to escape from Earth’s potential well?g) What should the launch velocity be it the satellite is required to escape from Earth’s potential well?

245.98 10EM kg 66.38 10Er m



a)What is the radius of the satellite’s orbit?

245.98 10EM kg 66.38 10Er m

Since we are given the period of the orbit is 24 hours. We will use the Period version of centripetal acceleration.

c GF F

2E S

S c

GM mm a

r

2E

c

GMa

r

2

2 2

4 EGMr

T r

23

24EGM T

r

2

324EGM T

r

2

211 2423

2

7

6.67 10 5.98 10 24 3600

4

4.23 10

N mkg h s

kg

m



b) What is the velocity of the satellites orbit?

245.98 10EM kg 66.38 10Er m

EGMv

r

2

11 242

7

6.67 10 5.98 10

4.23 10

N mkg

kg

m

4

3070

1.11 10

m

skm

h



c) What is the gravitational potential energy (universal) of the satellite when it is at rest on Earth’s surface?

245.98 10EM kg 66.38 10Er m

E sG

E

GM mU

r

211 24 2

2

6

10

6.67 10 5.98 10 5.00 10

6.38 10

3.13 10

N mkg kg

kg

m

J



d) What is the total energy of the satellite when it is in geosynchronous orbit?

245.98 10EM kg 66.38 10Er m

TE K U

21

2E sGM m

mvr

1

2E sGM m

r

92.36 10 J



e) How much work does the launch rocket do on the satellite to place it into orbit?

245.98 10EM kg 66.38 10Er m

f i

W E

E E

9 102.36 10 3.13 10J J

102.89 10 J



f) Once in orbit, how much additional energy would the satellite require to escape from Earth’s potential well?

245.98 10EM kg 66.38 10Er m

92.36 10 J

The negative of Total Energy to get the satellite

there.



g) What should the launch velocity be if the satellite is required to escape from Earth’s potential well?

245.98 10EM kg 66.38 10Er m

41.12 10m

s

2 EE

E

GMv

r

2

11 242

6

2 6.67 10 5.98 10

6.38 10

N mkg

kg

m


Suppose your hand moves up 0.50m while you are throwing a ball ( 0.145kg), which leaves your hand with an upward velocity of 20.0 m/s. Assuming your hand exerts a constant upward force on the ball, determine the magnitude of this force. Determine the speed of the ball 15.0 m above the point where it leaves your hand.

We will look at the energy diagrams for y1, y2, and y3 to guide us in solving this problem

Suppose your hand moves up 0.50m while you are throwing a ball (0.145kg), which leaves your hand with an upward velocity of 20.0 m/s. Assuming your hand exerts a constant upward force on the ball, determine the magnitude of this force. Determine the speed of the ball 15.0 m above the point where it leaves your hand.


The ball’s motion has two stages: while it is in contact with your hand and after it leaves your hand

To keep track of these stages, we let point 1 be where your hand first starts to move, and point 2 where the ball leaves your hand, and point 3 where the ball is 15.0m above point 2.

1

2

3



1

2

3The non gravitational force of your hand acts only between points 1 and 2. using a coordinate system with point 2 being at 0m and point 3 at 15.0m and point 1 at -0.50m, provides us with a easy application of conservative energies to solve this problem.

1 1 2 2

2 21 1 2 2

2

2 2

2 1 2 1

1 1

2 2

10 0.145 9.80 0.50 0.145 20.0 0.145 9.8 0.0

2

29.0 0 0 0.71

29.7

K hand K

hand

hand

hand K K

E U W E U

mv mgy W mv mgy

m m mJ kg m W kg kg m

s s s

W E E U U

J J J

J

The kinetic energy increases by 29.0J and the potential energy

increases by 0.71J, the change in total mechanical energy 29,7J is

due to the work by your hand.



1

2

3Since the force done by your hand was a constant force,, the Whand done by this force is equal to the force multiplied by the upward displacement (y2-y1)

2 1

2 1

29.7

0.5059

hand

hand

W F y y

WF

y y

J

mN



To find the speed at point 3, we note that the total mechanical energy between points 2 and 3 are conserved, Therefore we can use:EK2+U2=EK3+U3

1

2

3

2 2 3 3

3 2 2 3

2 23 2 2 3

23 2 2 3

2

2

1 1

2 2

2

20.0 2 9.8 0 15

10.3

K K

K K

E U E U

E E U U

mv mv mgy mgy

v v g y y

m mm m

s s

m

s

Documents

SPH4UI Gravitational Potential energy Mr. Burns. To explore how much energy a spacecraft needs to escape from Earth’s gravity, we must expand on the topic