1.01. space gravitational pe

fysbook SPACE Bryan Maher

Space

HSC Topic 1 – Focus 1 – Gravitational PE

created 13.75 billion years ago


1. The Earth has a gravitational field that exerts a force on objects both on it and around it

Students learn to:

Students:

define weight as the force on an object due to a gravitational field perform an investigation and gather

information to determine a value for acceleration due to gravity using pendulum motion or computer-assisted technology and identify reason for possible variations from the value 9.8 ms-2

gather secondary information to predict the value of acceleration due to gravity on other planets

analyse information using the expression:to determine the weight force for a body on Earth and for the same body on other planets

explain that a change in gravitational potential energy is related to work done

define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field

F mg

E p Gm1m2

r


Weight and the Gravitational FieldEvery massive body has an associated gravitational field surrounding it, extending out to infinity but weakening with distance. The field due to a body can be defined as the region of space surrounding it where other bodies will feel a force due to it.

Thus, if a second mass enters that field, it will experience a force of attraction -

- and, in turn, it will exert a force of attraction on the first mass.

Why?


This gravitational force, Fg, is the weakest of the four fundamental natural forces.

The gravitational field due to a body extends to infinity, so the gravitational force is infinite in range, although it becomes very weak at large distances as it is an inverse square law.

What does an “inverse square law” mean?

What does an “inverse square law” look like graphically?


The gravitational field can be visualised in terms of lines of force, or field lines – with the direction of the field lines indicating the direction of the gravitational force, and the relative spacing of field lines giving an indication of the gravitational field strength.

The gravitational field strength g is thus a vector, and the combination of the vectors at all points describes the gravitational field.

For a spherical object such as the planet Earth, the field lines are as shown, indicating a radially inward field which weakens with distance from the centre of the Earth.

Why do the field lines never cross?


Close to the Earth’s surface, the gravitationalfield is effectively uniform.

Note that a line (or more correctly, a surface) perpendicular to the field lines and joining places of equal gravitational field strength also represents a constant level of gravitational potential energy.

How does this link with what you know already about GPE?


The gravitational force acting on an object is defined as its weight, W.

The strength of the gravitational field is defined as the force per unit mass it exerts on a mass within the field.

That is, the size of the gravitational force acting on a mass m defines the strength of the gravitational field, g, at a point, by

g = Fg = W m m

The units of g are N/kg.

What else is defined as force per unit mass?


Notice that the strength of the gravitational field at a point does not depend on the size of the mass m placed in it – but only on the size and location of the masses which create the field.

Isaac Newton

I knew that – have a look at my equation for gravitational force.

Like .Comment .Share about 320 years ago


Calculation of W

Planet g at surfaceNkg-1

Weight at surface N

Mercury 3.78

Venus 8.60

Earth 9.78 978

Mars 3.72

Jupiter 22.9

Saturn 9.05

Uranus 7.77

Neptune 11.0


The gravitational field strength for a uniform spherical body of mass M is

given by

where G is the universal gravitational constant = 6.67 10-11 Nm2kg-2 and r is the distance from the centre of mass of the body.

g = GM r 2

Where does this equation come from?


Calculation of g

Planet Masskg

Diameterkm

g at surfaceNkg-1

Mercury 3.34 × 1023 4 880 3.78

Venus 4.87 × 1024 12 100 8.60

Earth 5.98 × 1024 12 800 9.78

Mars 6.40 × 1023 6 790 3.72

Jupiter 1.90 × 1027 143 000 22.9

Saturn 5.69 × 1026 120 000 9.05

Uranus 8.67 × 1025 51 800 7.77

Neptune 1.03 × 1025 49 500 11.0


Notice that for a freely falling object of mass m,

then a = Fg m

= mg m

= g

Fg

Albert Einstein

That is, the magnitude of the acceleration of a freely falling object is equal to the gravitational field strength at that point - and so...



For example…..1. What would a (very) accurate set of scales indicate as the weight of Liam at

New York (g = 9.803 ms-2) as compared to the equator (g = 9.780 ms-2)?2. What are some reasons for the variation in g at different points on the Earth’s

surface?3. How would you expect g at the North pole to be different compared to that at

the equator?4. What weight force would Phoebe experience on the surface of Jupiter

(g = 24.8 ms-2)? 5. If the Neptunian moon Triton has a mass of 2.14 1023kg and a radius of

1.35 106m, determine the intensity of the gravitational field at the surface and at an altitude of 100 km.

6. There exists between the Moon and the Earth a “parking space” for spacecraft where the gravitational field is effectively zero. This is known as the Langrangian point, and is shown as….

Questions SET 1.1


Lagrangian point

How does this come about?


ME = 5.98 ×1024kgRE = 6.37 × 106 m

MM = 7.36 ×1022kgRM = 1.74 × 106 m

3.82 × 108m

Lagrangian point

Solution


Gravitational Potential Energy The energy an object has as a result of its position in space relative to other massive objects is its gravitational potential energy. When work is done on an object to move it against a gravitational force, the gravitational potential energy of the object is increased.

Initial GPE

Final GPE (greater)


Gravitational Potential Energy

Work done against the field = Ep

= Fd

= mgh

Initial GPE

Final GPE (greater)

hF


Likewise, if the object moves freely under the influence of the gravitational field, its gravitational potential energy is decreased (and its kinetic energy is increased).

That is,

Work done by field = -EP = EK

Energy changes with height.xls

What if the mass moves horizontally ? Does the actual path

matter?


The previous analysis involved the special case where it was assumed the gravitational field was uniform over the distance the mass was moved.

As this is not necessarily the case, a better analysis uses the definition that the gravitational potential energy (EP) of an object of mass m1 at a

distance r from the centre of a mass of m2 is defined as the work done in

moving m1 from infinity to a distance r from the centre of m2.

This gives the result that

EP = - G m1m2

r

Isaac Newton

Solving problems like this was one of the reasons I invented calculus!




This equation assumes the gravitational potential energy to be zero at an infinite separation of the masses..

m1 m2

r = EP = 0

Thus, since a mass released at infinity will lose potential energy and gain kinetic energy as it accelerates under the influence of the gravitational field, it must have increasingly negative values for its gravitational potential energy.

m1 m2

r < EP decreases, so is < 0

Why?

So how can the change in GPE be positive?


EP

r

Plot of gravitational potential energy against distance from the centre of Earth.It is valid only for points beyond the radius of the Earth, rE

rE

Why?


EP

r

Since PE = KE = Work done

rE

So what is happening if, for example, a satellite is lifted up to a higher geostationary orbit?

rirf

GPEi

GPEf

KE = Work done by field =


If gravitational force per unit mass is plotted against distance from the centre of a planet….

Fg

m

rrE

9.8 Nkg-1

Why is this curve “upside down” compared to the previous one?

Is this curve just “upside down” compared to the previous one?


…the area under the curve represents the work done per unit mass either by or against the field in varying the distance of the mass from the centre of Earth.

rrE

Fg

m


0 5000 10000 15000 20000 25000 30000 35000 400000

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

Gravitational force on a 1000 kg satellite at varying distance from Earth's centre

Radius (km)

Fo

rce

(N

)


For example…1. Use the formula to determine the GPE of a 100kg object at the surface of

the Earth, and at a height of 1000m.2. What KE would result from the object falling from 1000m to the surface?3. Compare this to the value obtained using PE = mgh.4. Using the plot of Fg vs r from the previous slide, estimate the work needed

to lift a 1000kg satellite from an orbit of radius 10000km to one of 20000km.


For a body moving freely in a gravitational field, the total energy remains constant. Thus

r

mmGmvET

212

2

1

Hence, satellites in circular orbits have constant EK and EP , while those inelliptical orbits vary their EK and EP.

Johannes Kepler

I told you so….


Why?


For example…1. By considering the Law of Conservation of Energy, explain why the

sign for gravitational potential energy is negative.2. Explain the difference in kinetic energy for a satellite at aphelion

compared to perihelion.3. Calculate the change in potential energy if a 100 kg satellite is moved

from a height of 200 km above the Earth's surface to a height of 3400 km.

4. A 1kg particle is traveling radially in toward Earth at 10ms-1 at an elevation equal to the Earth’s radius. If air resistance is neglected, with what speed does the particle strike the Earth’s surface?

Questions SET 1.2


“A more modern view on this topic was presented by Albert Einstein. In a far more complex description, dealing with curved space, mass-energy tells space-time where to bend and vice versa. Obviously, the effects on everyday life are negligible. For the sake of completeness, it should be remarked that there are indeed observable relativistic effects, such as the trajectory of light being bent by the sun's mass. To summarise, Einstein's relativistic description of gravity is more accurate, far more complicated, of negligible effect on everyday life, and still incomplete!”


Determination of gMethod 1: Using computer assisted technology

By measuring values of displacement at different times, plot a curve of s/t vs t and determine a value for acceleration due to gravity.


Method 2: Using motion of a pendulum

By measuring values of period, T, and using

plot a curve of T vs l and determine a value for acceleration dueto gravity.

g

lT 2

1 period

Date post:	11-Jun-2015
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1.01. space gravitational pe

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