Topic 8 - Gravitational and Electric Fields

 

Key words

Gravity, radial, point mass, attractive

You are expected to:

·        recall and use the equation F = -Gm1m2

                                                                  r2

·        be familiar with the concept of gravitational field strength and the equations associated with it, g = F/m, g = -GM/r2

·        should know how to use the equation V = -GM/r. 

·        understand the graphical representation of g and V with r. 

·        be aware that the gravitational field strength, g = DV/Dr

·        be able to apply these ideas to satellites in orbit and the circular motion of planets.

 

Gravity Fields

A field is a region in which a force is felt.

 

Gravity is a very mysterious force.  Nobody knows why objects have this attractive force between them, even if they are far apart.  This attraction occurs for any object with mass, however small.  The force is very small, and always attractive.  We never get a repulsive gravitational force.

 

Newton found that all objects accelerate towards the Earth at 9.81 m/s2.  He also worked out the moon has an acceleration towards the Earth of 2.72 m/s2.  Therefore:

 

            Acceleration of Moon = 2.72 × 10-3 m/s2 =   1   =   1

            Acceleration of apple          9.81 m/s2      3610   (60.1)2

 

He also worked out the distances between the moon and the centre of the Earth and the distance between the apple and the centre of the Earth.  Their ratio also turned out to be 60.1:1.  This formed the basis of Newton’s Inverse Square Law of Gravitation.

 

Every particle of matter in the Universe attracts every other particle with a gravitational force that is proportional to the products of the masses and inversely proportional to the square of the distance between them.

 

We can write this in code form:

 

We can now add the constant of proportionality:

 

                                               

Learn this

 

The Gravitational Constant, G, has a value of 6.67 ´ 10 –11 N m2 kg-2.  The minus sign tells us that gravity is an attractive force.  In gravity problems, we always assume that the masses are point masses.

 

Question 1

What is the force of attraction between two giant ships each of mass 100 000 tonnes, 50 m apart? 

ANSWER

 

Gravity is a very small force, but it exists between all objects.  There is a tiny gravitational attraction between you and the student sitting next to you in your physics lesson.  You can't feel it because it's negligible.  The only reason we feel gravity is that the Earth is a very large object.  Remember that as well as the Earth attracting us towards it, we are attracting the Earth towards us.  The tiny attraction has important implications for dust particles coming together in space to form stars and planets.

 

 

Gravitational Field Strength

We are all familiar with the magnetic field of a bar magnet.  The field lines show up as areas of attraction and repulsion at the poles and the concentration of field lines at any point.  The closer we are to the magnet, the stronger the lines of force.

 

We can do a similar exercise with a gravity field; only this time there are attractive forces involved and no repulsive forces. 

The Earth, in common with many other planets, is very nearly a perfect sphere (relatively smoother than a billiards ball).  Its gravity field is radial with the pull being directly towards the centre.  The closer in we get, the stronger the pull.

Question 2

How can you tell from the diagram that the gravity field gets stronger?

ANSWER

 

The concentration of gravitational field lines is an indication of the gravitational field strength at any point, which is formally defined as:

 

The gravitational force per unit mass at that point.

 

So we can write that statement in code as:

 

            g = gravitational force = F  [Units – newtons per kilogram (N/kg)]

                        mass                 m

 

 

You will have met the expression F/m in the context of a = F/m, so it doesn’t take a genius to see that gravitational field strength is the same thing as acceleration.  A gravitational field strength of 9.81 N/kg causes an acceleration of 9.81 m/s2.

 

From our derivation of gravitational field strength and Newton’s Law of Gravitation, we can derive an equation to tell us the value of the field strength at a distance r from the centre of the Earth, mass M.

 

            F = -GmM                  Þ g = F/m  Þ g = -GmM = -GM

                     r2                                                     m r2       r2

 

 

 

Gravitational Field strength is a vector because it has a direction.  Remember always to take the radius of the Earth into account.

 

Worked Example

Assuming the Earth to be a perfect sphere, find its mean density from the following data:

  g = 9.81 N/kg; r = 6.37 ´ 106 m; G = 6.67 ´ 10 –11 Nm2 kg-2.

Work out the mass using g = - GM

       r2

 

            M = gr2 = 9.81 N/kg ´ (6.37 ´ 106 m)2 = 5.96 ´ 1024 kg

                  G             6.67 ´ 10-11 Nm2 kg-2

Now work out the volume of the Earth using V = 4/3 pr3:

 

V = 4 ´ p ´ (6.37 ´ 106 m)3 = 1.083 ´ 1021 m3

                      3

Now use density = mass ¸ volume to get the answer:

 

Density = 5.96 ´ 1024 kg = 5500 kg m-3

             1.083 ´ 1021 m3

  

Question 3

Jupiter has a mass of 1.91 × 1027 kg and its radius is 7.14 × 107 m.  What is its gravity constant?  What is the acceleration due to gravity?

ANSWER

         

If we draw a graph of the gravity field strength going to the centre of the Earth, we would not find a simple inverse square relationship, as if the Earth were a point mass.  Instead we would see a more complex variation.  The Earth is not a uniform sphere, but has a dense core and a less dense crust. The value of g can be calculated from seismic surveys and the graph above shows its variation with distance.  Note that the graph is drawn as a negative function as the field is attractive.  You may see it drawn as a positive function.

 

 

At the centre of the Earth, g = 0, because matter will be pulled equally in all directions so that the overall force is zero.

 

A bear-trap

 

When we work out g for distances away from the Earth's surface, we MUST add on the radius of the Earth.  Suppose a satellite were flying 1.0 × 106 m above the Earth's surface the distance r would NOT be 1.0 × 106 m but (1.0 × 106 m + 6.37 ´ 106 m) = 7.37 × 106 m.

 

Question 4

A satellite is 4000 km above the Earth's surface.  What is the acceleration due to gravity at this point?  Mass of Earth = 6.0 × 1024 kg

ANSWER

 

 

Gravitational Potential

The gravitational potential of a point is defined as:

 

The work done on a unit mass in moving it to that point from a point remote from all other masses.

 

In other words, this means the work done to move a unit mass from infinity to the point under consideration.

 

The zero point for gravitational potential is at infinity, so as we are moving towards the Earth, we are getting work out of the system.  Therefore gravitational potential is negative.  If we moved the object away to infinity, we would have to do a job of work on the object.  We work out gravitational potential with the formula:

 

 

 

[G = 6.67 ´ 10-11 Nm2 kg-2; M = mass of Earth or other planet; r = distance from centre.]

 

If we go back to our graph of gravitational field strength against distance we get:

 

 

We see that the gravitational potential is the area under the graph:

 

Question 5

The asteroid Ceres has a diameter of 785 km and a mass of 1.0 × 1020 kg.  Calculate:

(a)    the strength of the gravitational field;

(b)   the gravitational potential at the surface.

ANSWER

                  

 

If we plot V against r, we get the following graph:

 

 

From this graph we can see that the gradient gives us the gravitational field strength, g.  So we can say that:

 

 

 

We can go on to use the idea of gravitational potential to find an expression for the potential energy.  We have defined potential as energy per unit mass, so we can work out the total energy for any mass by multiplying the potential by the mass:

 

                                               

 

Another bear trap

Ep = mgDh is only true when we are very close to the Earth’s surface.  So do NOT use it for objects out in space.  Instead use

 

Worked Example

A satellite of mass 200 kg is to be moved from an orbit of 200 km above the Earth's surface to a an orbit of 400 km.  What work needs to be done?

(Radius of Earth = 6.37 ´ 106 m; mass of Earth = 6.0 × 1024 kg)

We need to calculate the gravitational potential energy at 200 km:

200 km = 0.2 × 106 m Þ r = 6.57 × 106 m

Ep = - 6.67 ´ 10-11 Nm2 kg-2 × 6.0 × 1024 kg  × 200 kg = -1.22 × 1010 J

                                    6.57 × 106 m                                      

 

Now do the same for the 400 km:

 

400 km = 0.4 × 106 m Þ r = 7.01 × 106 m

Ep = - 6.67 ´ 10-11 Nm2 kg-2 × 6.0 × 1024 kg  × 200 kg = -1.18 × 1010 J

                                            6.77 × 106 m        

 

Now we can work out the job of work we have to do to shift the satellite:

Work done = -1.18 × 1010 J - -1.22 × 1010 J = +0.04 × 1010 J =  +4.0 × 108 J

 

The plus sign tells us that a job of work has to be done.

 

 

Question 6

Some deep space cosmonauts land on a small planet.  They know that the mass of their craft is 500 tonnes and that to stop, they had to use 6.0 × 1011 J of energy.  They also worked out that the radius of the planet is 1500 km. 

(a)     Explain why the potential energy is - 6.0 × 1011 J.  Why is the sign negative?

(b)    What is the mass of the planet? 

(c)     What is its density? 

ANSWER

 

 

Motion of Masses in Gravitational Fields

Newton’s Laws of Gravitation can be used to explain the motion of planets and stars.  Much of modern space exploration uses the three hundred-year-old model.  Orbiting satellites are NOT doing gravity defiance acts; instead they are actually falling in a curved path towards the Earth all the time.  However they have a sufficient forwards velocity to miss the Earth all the time.  Since the gravity field is radial, the force acts at 90o to the direction of travel all the time.  Therefore the path is circular.  If we stopped gravity, the satellite would fly off tangentially into space in a straight line.  If we stopped the satellite, it would fall straight back to Earth.

 

 

For a satellite to be in a particular orbit, a particular velocity is required or a given distance.  Some satellites are placed so that they go in an easterly direction, completing one orbit each day.  They remain above one given point on the Earth’s surface, so are called geostationary.  This kind of satellite orbit is used in telecommunications.  Other satellites move in a polar orbit so that they can perform sweeps of the surface.  Spy satellites use a polar orbit.

 

When considering the motion of satellites in orbit, you have to know the rules of simple circular motion.  Click HERE to go back to Topic 1 to revise these.

 

Useful formulae include:

 

Worked Example

A communications satellite is to be placed in a circular geostationary orbit.  What must its height and speed be?

This question seems to be remarkably lacking in information, but there is an answer.

 

Use Newton’s Laws of Gravitation to solve this.  There is a single force acting on the satellite, gravitational attraction, so the satellite is acceleration all the time towards the centre of the Earth.

We need the satellite to be traveling at a sufficient forwards velocity so that it completes ONE orbit every 24 hours.  We need to work out the angular velocity before we can work out the linear speed.

w = 2pf.  We need to work out f.  f = ____1_____ = 1.16 ´ 10-5 Hz

                                                      24 ´ 60 ´ 60

 

Þ w = 2 ´ p ´ 1.16 ´ 10-5 Hz = 7.27 ´ 10-5 rad/s.

Now we need to consider the centripetal acceleration:

 

a = w2r.  We also know that the acceleration is also given by g = -GM/r2

 

Þ w2r = GM/r2 [We will ignore the minus sign.]

Þ r3 = -GM = 6.67 ´ 10-11 N m2 kg-1 ´ 5.98 ´ 1024 kg = 7.55 ´ 1022 m3

                                    (7.27 ´ 10-5 Hz)2

 

Þ r = 3Ö 7.55 ´ 1022 m3 = 4.24 × 107 m

Now we can work out the forward velocity:

 

v = wr = 7.27 ´ 10-5 rad/s ´ 4.24 ´ 107 m = 3100 m/s.

 

 

Question 7

The planet Mars has a diameter of 6800 km.  A satellite is in orbit 5000 km above the planet's surface travelling at a speed of 7100 m/s. 

(a)    How long does it take to orbit? 

(b)   What is the centripetal acceleration at this speed?

(c)    What is the acceleration due to gravity at this distance?

(d)   Will it remain in that orbit?  Mass of Mars = 6.42 × 1023 kg  

ANSWER

 

Summary

Force of gravity:

                                                   F = -G Mm

                                                              r2

Gravitational Field Strength:

Force per unit mass

                                                      g = -G M

                                                              r2

Gravitational Potential:

Energy per unit mass

                                                  V = -G M

                                                            r

 

Gravitational Potential Energy:

Energy needed to bring an object from infinity to a second object.

 

                                                  Ep = -G Mm

                                                              r

 

 

Now try the Topic Quiz Home Physics A2 Module 4