30

Physics Factsheet January 2002

Number 30

Gravity and potential Gravity acts on all bodies that have mass. The gravitational force between two bodies is: • always attractive;

• • • •

The gravitational force exerted by the earth on a mass is the weight of the mass, which you will be familiar with from the formula W = mg, where g is the acceleration due to gravity.

along the line joining the centers of mass of the two bodies;

This can be used to derive a relationship between g and G:

of the same magnitude on both bodies, but acting in opposite directions; Consider a mass m at the earth’s surface.

proportional to the product of the two masses F ∝ m1 m2; inversely proportional to the square of their seperation F ∝ 12 . r

W = mg

m

rE m1

F

F

mE

m2

We can write two expressions for the gravitational force acting on this mass due to the earth by using W = mg and Newton's Law of Gravitation. Equating these gives:

r

Newton’s Law of Gravitation states that the force between the attracting masses is : Gm 1 m 2 where G is the universal gravitational constant. F= (G = 6.67 × 10−11 Nm2kg−2). r2

mg =

where mE = mass of the Earth rE = radius of the Earth

Gm mE rE 2

Cancelling m gives: g=

Exam Hint: - Ensure you measure the seperation, r, between the centres of mass of the two bodies, not between their surfaces.

GmE rE2

This approach can be used to calculate the acceleration due to gravity on any planet, by substituting the mass and radius of that planet for the mass and radius of the earth. eg. Jupiter has mass 1.90 × 10 kg and radius is 6.99 × 10 m. So the acceleration due to gravity on Jupiter is GmJ /rJ 2 = 26ms-2 27

Worked Example. Two heavyweight boxers stand 1 metre apart. If they each have a mass of 80 kg, what is the gravitational force of attraction between them?

Gravitational field A field is a region where we can detect a force. To find the gravitational field strength at a point, we place a mass m at the point and measure the (gravitational) force F.

m1 = m2 = 80 kg. F=

6.67 × 10−11 × 80 × 80 12

7

newtons.

The gravitational field strength is defined to be the force per kg.

This is about 4.3 × 10−7 N, which is extremely small.

This ratio is also called the acceleration due to gravity g. Since g = F/m, the field strength may be quoted in Nkg−1, or ms−2.

Worked Example. Calculate the gravitational force of attraction between one boxer (mass 80 kg ) and the Earth. Mass of Earth mE = 5.98 × 1024 kg. Radius of Earth rE = 6.36 × 106 m.

Field strength above the earth's surface Consider a body of mass m at a point P a distance R from the centre of the Earth.

This calculation is fairly straightforward but you should be careful with powers of ten, particularly when squaring distances in the denominator.

mE

F

F

m

P

The calculation now looks like this: Gm1m2 F= r2 6.67 × 10−11 × 80 × 5.98 ×1024 newtons F= ( 6.36 × 106)2 F = 7.89 × 102 N

R GmE m R2 Divide both sides by m gives the gravitation field strength g:

The force F is given by F =

1

Gm F Gravitational field strength = m = g = 2 E - an inverse square law. R

Physics Factsheet

Gravity and potential

Weightlessness is a term often used. However, as can be seen from the graphs and formula, there is always a force between the Earth and any mass and so the mass always has a weight. When you see pictures of orbiting astronauts it is better to describe them as apparently weightless.

Field strength below the earth's surface For points below the earth's surface, the situation is slightly more complicated:

Field Lines Field lines are used to show the presence of a gravitational field. • The direction of the gravitational field lines shows the direction of the force acting on a mass in the field. • The spacing of the lines indicates the strength (or intensity) of the field. Lines close together = strong field; lines wide apart = weak field.

R P

The diagram below shows the gravitational field of a spherical mass such as the Earth. A field like this is called a radial field.

The field strength at point P is due only to the shaded sphere, of radius R, rather than the whole mass of the earth. If we assume the density of the earth is approximately constant, the mass of the shaded sphere can be expressed as a fraction of the mass of the Earth by using: mass of smaller sphere mass of Earth mS mE

So:

= =

volume of smaller sphere volume of Earth 4πR3/3 4πrE3/3

mS = mE R3 / rE3

So:

At increasing distance from the mass, the spacing increases showing that the field strength decreases - it follows the inverse square law detailed above.

The gravitational field strength will therefore be: GmE R3 / rE3 R2 GmE R = r3 E

Neutral point.

g=

For any two bodies there is a point P in between them where the pull of one body to the right is balanced by the pull of the other body to the left. This is called a neutral point. A mass placed at P would be in equilibrium.

So below the earth's surface, g is directly proportional to distance from the centre.

Consider the planet Jupiter and its moon Ganymede: Jupiter

The graph below shows the variation in gravitational field strength (g) with distance from the centre of the Earth. Note that the value of 9.81 that we often use is only correct at or near to the Earth’s surface - but even at the top of mount Everest, the reduction is less than 0.4%

Ganymede

Ganymede's orbital radius:1.06 × 109 m mG: 1.46 × 1023 kg mJ : 1.90 × 1027 kg.

P x

y

g/Ng-1 Since the gravitational force due to Jupiter is equal to that due to Ganymede at point P, we have

9.81

10

GmJ Gm G = x2 y2

8 6

g∝

Substituting in and rearranging gives: y2 ( 1.90 × 1027 ) = x2(1.46 × 1023) Square-rooting gives: 4.36 × 1013 y = 3.82 × 1011x So 114y = x

1 R2

4 2

Since x + y = Ganymede's orbital radius, we have 115y = 1.06 × 109, giving y = 9.2 × 106 m

Earth 0

rE

2rE

3rE

4rE

Worked Example (a) Calculate the acceleration due to gravity at a height of 50 km above the surface of the earth. Mass of Earth mE = 5.98 × 1024 kg. Radius of Earth rE = 6.36 × 106 m. G = 6.67 × 10−11 Nm2kg−2

(b) Calculate the height above the earth at which g is a quarter of its value on the surface of the earth. We need to find the value of R when g = GmE/rE2 Surface value of g = GmE/rE2 At distance R from the center, we have g = GmE/R2 So we need to solve GmE/R2 = GmE/4rE2 Hence R2 = 4rE2 So R = 2rE So the height above the earth's surface is rE = 6.36 × 106 m

Use g = GmE/R2 Note that R = radius of earth + 50km = 6.41 × 106 m So g = 6.67× 10−11 × 5.98 × 1024 /(6.41 × 106)2 g = 9.71 ms-2

2

F

Physics Factsheet

Gravity and potential Circular motion

There are two important things to note here: • We are only finding the increase in potential energy, we do not know the potential energy at a or b. • We have assumed that g is constant between the levels a and b.

Circular motion is covered fully in Factsheet 19, but to understand this section you just need to know that for a mass moving in a circle with constant speed: • Angular speed ω (rad s-1), linear speed v (ms-1) and radius of circular path r (m) are related by the equation v = ωr • The period of the motion (the time taken for one complete rotation) is given by 2π/ω • There must be a force directed towards the centre of the circle (the centripetal force) • The magnitude of the centripetal force is mω2r

The assumption that g is constant is valid close to the Earth's surface, but for positions further away from the earth, another approach is needed. mE m R2

F=G mE

R=a

m m

R=b

Consider the moon orbiting the Earth: R F

mE

mM

Suppose we wish to move the mass between the two points from where R = a to R = b and find the work done for this displacement. Because the force F is varying we cannot use the equation work = F× d.

R

Fortunately there is a simple way round the problem. First we find the work done in taking the mass from the point where R = a to infinity. For this displacement, the work done is GmEm a

The force F which is necessary for circular motion of moons around planets, or planets around the sun, is provided by gravity.

Since the centripetal force is supplied by gravity, we have: GmMmE = mM ω2 R R2 GmE = ω R 2

Hence:

NB: This equation only works in a radial field. To take the mass from the point where R = b out to infinity would require work of GmEm to be done. b So the change in potential energy between the two points is

3

Planetary system - Kepler's Laws Applying the same principle as above to any of the planets in the solar system orbiting the sun, we obtain: GmS = ω2 R3 where mS is the mass of the sun, and R is the orbital radius of the planet.

GmEm − a

GmEm b

= GmEm

(1a

−

1 b

)

Worked Example The Earth has a mass mE = 6.0 × 1024 kg and radius rE = 6.36 × 106 m. Find the gain in p.e. of a 1 kg mass when it is lifted from the Earth’s surface to a height of i) h = rE and ii) h = 2rE. iii)h = ∞ (infinity) Take G = 6.67 × 10−11 Nm2kg−2

But since the orbital period, T = 2π/ω, we have ω = 2π/T:

()

2 GmS = 2π R3 T

Rearranging: GmST2 = 4π2R3 This gives us: T 2 ∝ R3 This result is Kepler's Third Law of Planetary Motion.

i) Using the above, we have a =rE and b = 2rE ∆ (pe) = G mEm 1 − 1 . rE 2rE −11 24 1 ∆ (pe) = 6.67 × 10 × 6 × 10 × 1 × joules 2 × 6.36 × 106

[

]

[

NB: This derivation relies upon approximating the planets' orbits by circles, although they actually move in ellipses around the sun. However the result still holds true despite this.

]

∆ (pe) = 3.15 × 10 J 7

[ r1 − 3r1 ]. 2 × 1 × [3 × 6.36 × 10]

ii) Again a = rE , but b = 3rE . ∆ (pe) = G mEm

E

Gravitational Potential Energy

−11

∆ (pe) = 6.67 × 10 × 6 × 10 7 ∆ (pe) = 4.19 × 10 J

Consider a mass m in the Earth’s gravitational field. b

24

E

6

iii) Again a = rE , but now, b = ∞. ∆(pe) = GmEm 1 − 1 rE ∞ Note that ∞1 = 0 and so, ∆ (pe) = 6.67 × 10−11 × 6 × 1024 × 1 × 6.36 1× 106 ∆ (pe) = 6.3 × 10−7 J or 63 MJ

[

F m

∆h

[

].

]

a There is an important idea contained in this last part. The 1 kg mass has been taken to ‘infinity’ but the work done or gain in pe has increased by a finite amount ( 6.3 × 107 J).

mg To raise the mass from level a to b, the force F will do work F × ∆h on the mass, which increases the potential energy of the mass by this amount Since F = mg , the increase in potential energy is given by ∆(p.e.) = mg∆h.

3

Physics Factsheet

Gravity and potential

Lines of equipotential These are lines which surround a body, all points on the line being at the same potential. They are shown as dotted lines in the diagram below.

Defining Gravitational Potential Energy To define the actual potential energy, rather than working out differences in it, it is first necessary to decide at which point a body will have zero potential energy. At first sight it may seem strange but it has been agreed that this point shall be at ‘infinity’.

p

b

Lines of equipotential

q Since moving a point from the earth's surface to infinity increases its potential energy (as seen in the previous example), if the potential energy is zero at infinity, it must be negative elsewhere.

a

a

Gravitational potential energy of a mass at a point in a gravitational field is the work done in bringing the mass from infinity to that point. Gravitational Potential (V)

Lines of Field

Gravitational potential (V) is the gravitational potential energy per unit mass. GmE (Jkg−1) V=− R R = distance from centre of earth (R >rE)

Note that the lines of equipotential are perpendicular to the field lines. If you move along the curved path from a to b then the displacement is always perpendicular to the direction of the force. This means no work is done and there is no gain (or loss) in potential energy; hence the curve through ab is a line of equipotential.

To find the potential energy of a body at a given point, multiply the potential at the point by the mass of the body.

Worked Example A piece of space junk of mass 10 kg detaches itself from a spacecraft at p and falls freely to point q. Find the kinetic energy gained by the 10 kg mass in falling to Earth. ( rp = 30 × 106m, rq = 12 × 106m mE = 6 × 1024kg)

The diagram below shows how the potential varies with the distance from the earth’s centre. V/107 Jkg−1 Earth

5rE

0

Since the mass falls freely from p to q, the gain in kinetic energy is equal to the loss in potential energy. The potential energies at points p and q are m × Vp and m × Vq. where. V = − GmE R R

10rE 15rE R

−2 −4

V∝

Gain in ke = pe ( p ) – pe ( q )

1 R

= ( mVp ) – ( mVq )

−6

= m (Vp− Vq) =m

(− Gmr ) − (−Gmr ) E

E

p

q

[

]

Typical Exam Question.

= mGmE r1 − r1 . q p

(i)

= 10 × 6.67 × 10

(ii)

−11

Use Newton’s law of gravitational to find the units for G GmE −m× Hence show that the expression R does have the unit joule.

× 6 × 10

24

[ 12 ×1 10

6

−

]

1 6 30 × 10 .

= 2.00 × 10 J 8

Escape Velocity Answer: mm (i) Newton’s law: F = G r1 2 2

(ii)

re-arrange

The escape velocity is the velocity a mass must have to escape from the Earth ( or any celestial body) and not return. The mass must not be driven by motors of any kind, but have a velocity imparted to it that will carry it completely away from the Earth ie to infinity. As the mass gets further from the Earth, its potential energy increases. At infinity, the potential energy has increased to zero. When the mass has its escape velocity u, it must have sufficient kinetic energy to match the necessary gain in potential energy in taking it to infinity. Let u = the escape velocity for the Earth. We use conservation of energy for a mass m. Loss in kinetic energy = Gain in potential energy.

Fr2 G= mm 1 2

Substituting the known units, we obtain the units for G : 2 [G] = Nm2 or Nm2kg−2 kg Gm pe = −m × R E . Substitute the units: − kg × (Nm2kg−2) × kg [pe] = m = (−)Nm.

½ mu2 = pe(∞) − pe(rE) 2GmEm ½ mu2 = 0 − − rE

Since work (or energy) is force × displacement it follows that Nm is the same as joules.

[

u = 2GmEm = rE

4

]

2 × 6.67 × 10−11 × 6 × 1024 = 11 × 103 ms−1 6.36 × 106

Physics Factsheet

Gravity and potential

Exam Workshop a) Write down an expression for the gravitational potential at a distance R from the centre of a planet of mass M. Hence write down an expression for the potential energy of a mass m at a height h above the Earth’s surface ; radius of Earth = RE , mass Earth = ME . [3] GM Potential V = Potential energy = mgh. ! 1/3 R

e) Indicate on the graph the quantity that represents the energy E.

Force on satellite /kN

10

Minus sign omitted. Has not understood how to find potential energy from potential . Formula quoted only valid when h « RE . In this case h = RE . b) An amount of energy E is transferred to a satellite of mass m when launched by a certain rocket from the surface of the Earth. If the vertical height reached by the satellite is h, obtain an equation relating h to E and the other terms. [2] GME h. ! 0/2 RE 2 A fatal error has been carried forward and gains no credit. Substituting GME for g = is meaningless as this expression is only true at the Earth's RE2 surface. This candidate has relied too much on quoting formula without understanding the physics.

6 4

!

2 ! RE 2RE 3RE 4RE Distance from Earth centre

This answer is correct and gains full credit. It is testing how to find the work done by a force which varies with distance. This is a general result and is used elsewhere in physics.

Examiner’s answers

[2]

GM E ! R potential energy of mass m = m × V, at height h, R = R + h ! a) potential V = −

0/2

E

d) The graph below shows how the force acting on the satellite varies with distance from Earth’s centre.

and so p.e. = − m ×

GM E

(R E + h )

!

b) Energy E is transferred so by conservation, ! !  GM E   GM E   = − m  E +  − m R E   (R E + h )  

10 Force on satellite /kN

8

0

E = mgh = m

GMEm c) Show that if = RE, the energy E is given by E = 2RE GME Putting RE = h, E=m ! RE

8 6

 GM E c) When h = R we get E +  − m E RE 

4 2

E= m

0

RE 2RE 3RE 4RE Distance from Earth centre

!   GM E  = − m   2RE  

d) As above, force follows inverse square law:- F = G 1 1 Doubling distance gives   = of force. ! 4 2

The force follows Newton's Law of Gravity.

10 = 2.5kN ! 4 e) Shade area under graph from RE to 2RE ! !

2

Force when h = RE

Mm ! R2

1 Mm G 2 4 R ! (2R ) The force is reduced to a quarter, so force at height R is:

When h = R, the distance is doubled so F = G

 ⇒  

GM E! 2R E

Write the equation that shows how this force can be calculated and given that the force at the Earth’s surface is 10 kN, calculate the force on the satellite at a height h = RE. [3]

F =G

Mm

2

[2]

=

10kN = 2.5kN ! 4

This answer is correct and gains full credit. It is testing the inverse square law.

5

mM ! R2

Physics Factsheet

Gravity and potential

e) i) Write down an expression for the gravitational potential at a distance R from the centre of a planet of mass M. [2]

Typical Exam Question 1) i) Explain the term ‘gravitational potential’ as applied to a point in the Earth’s gravitational field. ii) The gravitational potential at the Earth’s surface is −63 MJ kg-1 ; taking the Earth to be a sphere of radius R what are the potentials at points A and B at heights R and 3R respectively above the Earth’s surface? iii) A 10 kg mass is moved from A to B. Does its potential energy increase or decrease and by how much ? iv) A 1 kg mass at the Earth’s surface is taken to infinity. What is the minimum amount of energy to achieve this ?

ii) Using the graph below find the potential energy of a 10 kg mass at 4×104m from the planet’s centre (point X). [2] V/kJ kg-1

· -20

ii) At the Earth’s surface radius R, the potential is given by GM V=− = −63MJ kg-1 R and the potentials at A and B are given by GNM ! GM ! VA = − and VB = − and so 4R 2R

· X

8

6

·

·

R/I

·

·

Y

·

iii) Find the work done in moving the 10 kg mass from point X to point Y (6×104m from planet’s centre). [2] iv) Write down the gravitational potential at point X and hence deduce the gravitational field strength (intensity) at point X. What force acts on the 10kg mass? [2 + 1]

VA = −

2 [4]

a) Define the term ‘gravitational potential’ at a point in a radial gravitational field. b) Use this definition to calculate the ‘escape velocity’ for the Earth with mass ME = 6×1024 kg and radius RE = 6.36×106 m.

iii) The potential energy is m × V joules. So, for the 10 kgmass at points A and B, we have: p.e. (A) = 10 × (31.5) MJ = −315 MJ p.e.(B) = 10 × (−15.75 ) MJ = −157.5 MJ !The potential energy at B is less negative than at A, hence in moving from ! A to B, the potential energy increase ! [3] iv)

4

-10

i) The gravitational potential at a point is the work done per kilogram!by an external force in moving a mass from the point to infinity!(or zero potential) Or, alternatively, the work done per kilogram by the field as the mass ‘falls’ from infinity to the point. [2]

62 63 MJ kg-1 and VB = − MJ kg-1 2 4 ! ! VA = − 31.5MJ kg-1 and VB = −15.75 MJ kg-1

2

0

c) A 1kg mass is to be catapulted from the Earth’s surface so that it will not return. How much energy must be given to the mass? What is this energy in kilowatt hours and how much would it cost (at the domestic rate of 5.5pence / kWh) to expel this mass?

The potential at infinity is zero, hence the potential energy of any mass at infinity is zero.!The potential energy of 1 kg mass at Earth’s surface is –63 MJ.!To move it to infinity the potential energy must increase to zero. Hence, the minimum energy required is 63 MJ.! [3]

d) The planet Mercury is smaller than Earth with a mass MM = 3.24 ×1023 kg and radius RM = 2.34 × 106 m. Calculate the escape velocity for Mercury and, giving reasons, comment on the likelihood of Mercury retaining any atmosphere. Assume G = 6.67 × 10-11 Nm2kg-2 3. a) Write down Newton’s law of gravity for two attracting masses.

Questions (Where necessary, take G = 6.67 × 10 Nm kg ) 1 a) Explain the term ‘gravitational field’ -11

2

-2

b) Define the term ‘gravitational field strength’

b) Assuming the Earth to be a uniform sphere of mass M and radius R show that the acceleration of free fall at the Earth’s surface may be deduced from Newton’s law and is given by:

[2] [2]

g=

c) A communications satellite orbits the Earth in a circular path of radius 4.2 ×107 m. Calculate the gravitational field strength at a point in this orbit (Mass of earth =6×1024kg) [3 ]

c) Given that the mean density of the Earth is ρ, show that G=

d) The figure below shows a spherical planet with some equipotentials. -700 Jkg-1

3g 4πρR

d) Show that the units on both sides of this equation are consistent.

-1000 Jkg-1 -1500 Jkg-1

a

R2 GM

e) A spaceship hovering above the surface of Mars releases a mass at a height of 190m . This mass hits the surface of the planet at 38ms-1. Calculate a value for the acceleration of free fall at the planet’s surface. f) Given that Mars has a radius of 3.4 × 106 m, estimate a value for the density of the planet. Assume G = 6.67 × 10-11 Nm2kg-2

c b d

Calculate the gain in potential energy when a 10 kg mass is moved i) from a to b ii) from b to c iii) from d to c [1+3]

6

Physics Factsheet

Gravity and potential Answers

Typical Exam Question

1. a) Space around a mass where gravitational effect (force on second mass ) can be detected.

2) A moon of mass m is orbiting its planet of mass M in a circular orbit of radius R. i) Draw a diagram showing the forces acting on both bodies. How do these forces arise and are they equal in magnitude or not ? [2] ii) Which of the two forces you have drawn provides the centripetal force necessary for circular motion ? The moon orbits the planet in a time T. [1] iii) State Kepler’s third law of motion as applied to the moon [2] iv) Show how Kepler’s third law may be deduced from Newton’s law of universal gravitation. [3]

b) Gravitational field strength /intensity at a point defined as force F per unit mass at that point = m F GM 6.67× 10-11× 6× 1024 mM c) F = G 2 ⇒ m = R = (4.2× 107)2 R = 0.227 Nkg-1 (ms-2) d) pe = m ×V gain pe(ab) = (-10×1000) − (-10×1500) = 5000J gain pe(bc) = (-10×700) − (-10×1000) = 3000J gain pe (dc) = zero.

i) m

M

e) i) potential V = − GM R

F1

F2

! ! F1 and F2 are mutual gravitational attraction. F1 = F2 ii) centripetal force = F2 ! iii) Kepler’s third law : T2 ∝ R3 ! Mm G 2 = m × Rω2 ! R

ii) from graph, at X, potential = -11.5kJkg-1 ⇒ pe(x) = -11.5×10 kJ = -115 kJ (allow for sensible error in graph e.g. 11