Module 5: C18 - Gravitational Fields

Question 1

What is Gravity, Gravitational Force, and Gravitational Field Strength

Answer 1

Gravity:
Gravity is the acceleration of one body to another because of its mass. (Measured in g)

Gravitational Force:
An attractive force, measured in Newtons (W=mg)

Gravitational Field Strength:
(Same as gravity)
Gravitational field strength is the acceleration of one body to another because of its mass.

Question 2

What is Gravity

Answer 2

Gravity is a universal attraction between any two masses.

It is a relatively weak force, so we only tend to notice it when one of the masses is very large, like the Earth or a star, but it is also present on a tiny scale.

Question 3

What are Gravitational Fields and What Happens in them

Answer 3

All objects with mass create a gravitational field around them. This field extends all the way to infinitely, but gets weaker as the distance from the centre of mass of the object increases, becoming negligible at long distances.

Any other object with mass placed in a gravitational field will experience an attractive force towards the centre of mass of the object creating the field. For objects on earth, we call this gravitational attraction the objects weight.

Question 4

How can the Gravitational Field Strength equation be written

Answer 4

The gravitational field strength g at a point within a gravitational field is defined as the gravitational force exerted per unit mass on a small object placed at that point within the field.

This can be written as:

g = F/m

Question 5

How can Gravitational Field Patterns be shown

Answer 5

We can map the gravitational field pattern around an object with gravitational field lines. These lines don’t cross, and the arrow show the direction of the force on a mass at that point in the field.

Since gravitational force is always attractive, the direction of the gravitational field is always towards the centre of mass of the object producing the field. The field lines around a spherical mass, like a planet, form a radial field. The gravitational field strength decreases with distance from the centre of the mass (shown by the field lines getting further apart)

Question 6

What is a Uniform Gravitational Field

Answer 6

If the field lines are parallel and equidistant, the field is said to be a uniform gravitational field. In a uniform field, the gravitational field strength does not change. The gravitational field close to the surface of a planet is approximately uniform.

Question 7

What does Newton’s Law or Gravitation State?

Answer 7

According to Newton’s 3rd Law, the two objects must experience a force F of the same magnitude but in opposite directions.

Newton’s law of gravitation states the form between two point masses is:
- Directly proportional to the product of the masses, F∝Mm
-Inversely proportional to the square of their separation, F ∝ 1/r^2

Therefore,
F ∝ Mm / r^2

(The equation contains a negative sign to indicate that the force is attractive.)

Question 8

How does the attractive force F in Newton’s Law of Gravitation act in relation to distance

Answer 8

The attractive force F between objects decreases with distance in an inverse-square relationship (F∝1/r^2). Double the distance and the force between objects will decrease by a factor of four.

Question 9

Worked Example:
The Gravitational Force on an Orbiting Satellite

A satellite of mass 70.0kg orbits the Earth at a height of 10,100km above the surface. The mass of the Earth is 5.97x10^24kg and it has a radius 6370km. Calculate the magnitude of the gravitational force on the satellite due to the Earth.

Answer 9

Step 1: Determine the distance of the satellite from the centre of the Earth

r = 6370 + 10100 = 16470km

Step 2: Use the equation for Newton’s law of gravitation to calculate the size of the force on the satellite

F = GMm / r^2
= 6.67x10^-11 x 5.97x10^24 x 70.0 / (16470x10^3)^2
= 103N (3sf)

Question 10

In Newton’s Law of Gravitation Fgrav = -Gm1m2 / r^2,

1. How does doubling the mass affect the force?
2. How does doubling both masses affect the force
3. How does doubling the distance apart affect the force

Answer 10

1. Doubling either mass doubles the force
2. Doubling both masses quadruples the force
3. Doubling the distance apart quarters the force

Question 11

Example Question

Two identical lead spheres have a combined mass of 12kg. The density of lead is 11,400 kgm-3.

a) Find the radius of each sphere.
b) Calculate the gravitational force of attraction between the spheres when they are touching.

Answer 11

a) ρ = m/v
v = m/p = 6/1400 = 1/1900

V = 4/3πr^3
1/1900 = 4/3πr^3
r^3 = 1.26x10^-4
r = 0.05m

b) F = Gm1m2 / r^2
6.67x10^-11 x 6 x 6 / 0.1^2
F = 2.4x10^-7 N

Question 12

What is a Gravitational Field Line

Answer 12

A gravitational field line is a line that indicates the direction of the gravitational force that would act on a test mass placed in the field.

Question 13

What is a Test Mass

Answer 13

A ‘test mass’ is a mass that is small enough not to affect the shape of the field with its own gravity.

Question 14

Where does the Earth’s Gravitational Field act

Answer 14

The Earth’s gravitational field appears to act from it’s centre of mass.

The Earth’s gravitational field gets weaker with distance from the Earth, this is shown by the greater separation of field lines.

Question 15

Rules of Magnetic Field Lines

Answer 15

• Lines never cross
• Lines always have an arrow to show the direction of Force produced
• The more lines per area, the stronger the field

Question 16

How can you detect small changes in Gravity

Answer 16

You can detect very small changes in gravity using a gravimeter, which can be used to map tunnels, caves, changes in volcanic activity or even minerals for mining.

Question 17

Is Gravitational Field Strength a scalar or vector quantity and why

Answer 17

Gravitational field strength is a vector quantity and always points to the centre of mass of the object creating the gravitational field.

Question 18

What is a stronger gravitational field represented by

Answer 18

A stronger field is represented by field lines that are closer together.

Question 19

What is Newtons Law of Gravitation (+ equation)

Answer 19

F ∝ Mm / r^2

We can write this as an equation using the gravitational constant G. Therefore, an equation for Newton’s law of gravitation is:

F = -GMm / r^2

Question 20

Why is a minus sign required in the equation:

F = -GMm / r^2

Answer 20

A minus sign is also required to show that gravitational force is an attractive force.

Question 21

Example Question:

Calculate the gravitational field strength due to the Earth at the Moon.

Earth Mass = 6.0x10^24kg
Earth Radius = 6400km
Moon Orbital Radius = 400,000km
G = 6.672x10^-11 Nm^2kg^-2

Answer 21

g = -GM / r^2

g = (6.672x10^-11) x (6.0x10^24kg) / (400,000x10^3m)^2

g = 4.0x10^14 / 1.6x10^17

g at the Moon due to the Earth’s gravity
= 0.0025 Nkg^-1

Question 22

Example Question:
The Sun has a mean radius of 700Mm and a surface gravitational field strength of 270Nkg^-1. Determine the mass of the Sun and it’s mean density.

(Gravitational constant G = 6.67x10^-11 Nm^2kg^-2)

Answer 22

r = 700x10^9 m
g = 270 kgN^-

g = GM/r^2
M = gr^2/G

270 x (700x10^6)^2 / 6.67x10^-11 = 1.98x10^30

ρ = M/V
ρ = M / (4/3πr^3)
ρ = 1.98x10^30 / (4/3π x (700x10^6)
ρ = 1.4x10^3 kgm^-3

Question 23

The gravitational field strength on the surface of the Earth is 9.81Nkg^-1. A satellite in a geostationary orbit round the Earth experiences a gravitational field strength of 0.225Nkg^-1. Determine the orbital radius, r, of the satellite from the centre of the Earth in terms of the radius of the Earth Re.

Answer 23

Ge = 9.81Nkg^-1
= -GMe / Re^2 => -GMe = 9.81Re^2 (1)

Gs = 0.225Nkg^-1
= -GMe/Ro^2 => -GMe = 0.225Ro^2 (2)

9.81Re^2 = 0.225Ro^2
Ro^2 = 9.81/0.225 Re^2
Ro = √9.81/0.225 Re
Ro = 6.6 Re
Re = 6371 km

Question 24

What is a Uniform Gravitational Field (How does it work)

Answer 24

In a uniform gravitational field, the gravitational field strength does not change. Close to the surface of the Earth, g is fair constant, and so the gravitational field can be considered approximately uniform.

Question 25

What can we see from the equation g = -GM / r^2

Answer 25

From this equation we can see that in a radial field, the gravitational field strength at a point is:

• Directly proportional to the mass of the object creating the gravitational field (g∝M)
• Inversely proportional to the square of the distance from the centre of mass of the object (g∝1/r^2)

Question 26

Worked Example: g on the International Space Station

The radius of the Earth is 6370km and it has mass 5.97x10^24kg. At 75kg astronaut in the International Space Station (ISS) orbits at a height of 405km above the surface of the Earth. Calculate the magnitude of the gravitational field strength at this altitude and the magnitude of the weight of the astronaut.

Answer 26

Step 1: Determine the distance of the ISS from the centre of mass of the Earth.

r = 6370 + 405 = 6775km

Step 2: Use the equation for gravitational field strength in a radial field to calculate g at this altitude. (The minus sign is not required for the magnitude.)

g = GM/r^2 = 6.67x10^-11 x 5.97x10^24 / (6.775x10^6)^2 = 8.67 Nkg^-1

Step 3: The weight W of the astronaut is given by W = mg.

W = 75 x 8.67 = 650N

Question 27

Worked Example: g in the Himalayas

It is suggested that objects are ‘lighter’ on the top of mountains as they are further from the centre of mass of the Earth. Calculate the percentage difference between the gravitational field strength on the surface of the Earth and on top of Mount Everest and comment on the effect.

Radius of the Earth = 6370km
Mass of the Earth = 5.97x10^24kg
Height of Mount Everest = 8840m

Answer 27

Step 1: Use the equation for gravitational field strength in a radial field to calculate g at sea level and on top of Everest. (The minus sign is not required for the magnitude)

Gsl = GM/r^2
= (6.67x10^-11 x 5.97x10^24) / (6.370x10^6)^2
= 9.81Nkg^-1

Gev = GM/r^2
= (6.67x10^-11 x 5.97x10^24) / (6.378840x10^6)^2
= 9.79Nkg^-1

Step 2: Calculate the percentage difference in g
((9.81-9.79) / 9.81) x 100 = 0.20%

Objects will weight slightly less, but the difference is negligible to a climber.

Question 28

Define gravitational field strength at a point in space.

Answer 28

Gravitational field strength is the gravitational force per unit mass on a point within a particular gravitational field

Question 29

Show that the gravitational constant G has the unit Nm^2 kg^–2.

Answer 29

g = GM/r^2
Nkg^-1 = Gkg/m^2
G = Nkg^-2 m^2

Question 30

Calculate the magnitude of the gravitational force between two binary stars, each of mass 5.0x10^28 kg, with a separation of 8.0x10^12m

Answer 30

g = GM/r^2
g = 6.67x10^-11 x (5x10^28) / (8.0x10^12)^2
g = 5.21x10^8 Nkg^-1

Question 31

Example Question:

The planet Neptune has a mass of 1.0x10^26 kg and a radius of 2.2x10^7 m.
Calculate the surface gravitational field strength of Neptune

Answer 31

g = GM/r^2
g = 6.67x10^11 x 1.0x10^26 / (2.2x10^7)^2
g = 13.8 Nkg^-1

Question 32

Example Question:

Calculate the radius of Pluto, given its mass is 5.0x10^23 kg and its surface gravitational field strength has been estimated to be 4.0 Nkg^–1.

Answer 32

g = GM/r^2 => r = √GM/g

r = √(6.67x10^-11 x (5.0x10^23)) / 4

r = 2.89x10^6m

Question 33

Example Question:

A space probe of mass 1800kg is travelling from Earth to the planet Mars.
The space probe is midway between the planets. Use the data given to calculate:

a) the gravitational force on the space probe due to the Earth
b) the gravitational force on the space probe due to Mars
c) the acceleration of the probe due to the gravitational force acting on it.

Answer 33

F = GMm / r^2

a)
Fe = (6.67x10^-11 x 1800 x 6.0x10^24) / (3.9x10^10)^2
Fe = 4.736x10^-4

b)
Fm = (6.67x10^-11 x 1800 x 6.4x10^23) / (3.9x10^10)^2
Fm = 5.05x10^-5

4.736x10^-4 - 5.05x10^-5 = 4.23x10^-4N

F = ma
4.23x10^-4 = 1800a
2.35x10^-7 = a

Question 34

What is Kepler’s First Law

Answer 34

The orbit of a planet is an ellipse with the Sun at one of the two foci.

An ellipse is a ‘squashed’ or elongated circle, with two foci. The orbits of all the planets are elliptical.

(In most cases, the orbits have a low eccentricity, so their orbits are modelled as circles)

Question 35

What is Kepler’s Second Law

Answer 35

A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

As planets move on their elliptical orbit around the Sun, their speed is not constant. When a planet is closer to the Sun it moves faster.

Question 36

What is Kepler’s Third Law

Answer 36

The square of the orbital period T of a planet is directly proportional to the cube of its average distance r from the Sun.

This can be written as a relationship as T^2 ∝ r^3

Question 37

For an Object moving in a Circle, what is the resultant force (Centripetal Force)

Answer 37

F = mv^2 / r

Question 38

Equation for mass (of the larger planet) when you combine:

F = mv^2/r
and
F = GMm/r^2

Answer 38

mv^2 / r = GMm / r^2

M = v^2r / G

Question 39

What the point at which a planet is closest to the sun called and furthest from the sun called

Answer 39

• The point that a planet is closest to the sun is perihelion
• The furthest point is aphelion

Question 40

What is the semi-major axis and the semi-minor axis in an ellipse

Answer 40

The semi-major axis is the distance from the centre (halfway between the foci) and the widest part of the ellipse.

The semi-minor axis goes between the centre and the narrowest part.

Question 41

How does the distance between the foci of an eclipse affect how eccentric it is?

Answer 41

The further the foci are from each other, the more eccentric the ellipse is.

Question 42

What is an Orbital Period?

Answer 42

The period of the satellite’s orbit, T, is the time it takes for the satellite to make one complete circuit

Question 43

Equation for T^2 using Kepler’s 3rd Law

Answer 43

Using T = 2πr/v
T^2 = 4π^2r^2 / v^2

Using v^2 = GM/r
T^2 = 4π^2r^2 x r/GM

T^2 = (4π^2 / GM) r^3

Question 44

What does the Mathematical version of Kepler’s Law, T^2 ∝ r^3

Answer 44

• The ratio T^2 / r^3 is a constant and equal to 4π^2/GM
• The gradient of a graph of T^2 against r^3 must be equal to 4π^2 / GM

Question 45

What is a Satellite?

Answer 45

A satellite is a body that circles around another body in space. Moons, comets, asteroids, planets, and stars are examples of natural satellites.

Question 46

What might a star be called a satellite?

Answer 46

• A star could be a satellite of another star in a binary system
• A star could be described as a satellite if the galactic centre
• When a planet appears to orbit a star, both star and planet are actually orbiting their common centre of mass.

Question 47

What Artificial Satellites exist?

Answer 47

Our Moon is a natural satellite of the Earth. Artificial satellites include the International Space Station (ISS) and the Hubble Space Telescope (HST).

Question 48

What are Satellites used for?

Answer 48

• Communications: Satellite phones, TV, some types of satellite radio
• Military uses: reconnaissance
• Scientific research: both looking down into the Earth to monitor crops, pollution, vegetation, and so on, and looking outwards to study the Universe (including several famous examples like the Hubble Space Telescope)
• Weather and climate: predicting and monitoring the weather across the globe and monitoring long- term changes in climate
• Global positioning

Question 49

If a satellite orbits the earth at a lower altitude, how does its speed need to change to stay in orbit?

Answer 49

This means that if a satellite orbits the Earth at a lower altitude, it needs to travel faster to stay in orbit.

Question 50

What are Polar Orbits and why are they used?

Answer 50

● Low orbit around the Earth passing over North and South poles.

● Earth rotates underneath them as they orbit.

● Used for large-scale mapping and global weather monitoring.

Question 51

What are Geostationary Orbits and how are they used?

Answer 51

● Stay above the same point on Earth.
Geostationary Satellites
● Speed of orbit matches the Earth’s rotation, so orbit time is 24 hours.
● Used for communications, satellite TV, weather forecasting, intelligence, global positioning system (GPS).

Question 52

What types of orbits are used for satellites

Answer 52

• Polar Orbit
• Geostationary Orbit
• Low Earth Orbit

Question 53

Worked Example: Height of Geostationary Satellites

The radius of the Earth is 6370km, and it has mass 5.97x10^24kg. Calculate the altitude (height above the ground) of geostationary satellites above the equator.

Answer 53

Step1: Rearrange the equation for Kepler’s third law to make r the subject.
T^2 = (4π^2 / GM)r^3
r = ∛(GMT^2 / 4π^2)

Step 2: The period of a geostationary satellite is 24 hours = 86400s. Substitute this and the other values in to calculate r.
r = ∛(6.67x10^-11 x 5.97x10^24 x 86400^2 / 4π^2)

Step 3:
Subtract the radius of the Earth to obtain the altitude of the satellite.
Altitude = 4.22x10^7 - 6.37x10^6 = 36x10^6m

Question 54

Equation showing how fast something travels in orbit

Answer 54

Fgrav = GMm/r^2
Fcentripetal = mv^2/r

GMm/r^2 = mv^2/r

v = √GM/r

Question 55

What is ΔV a change in?

Answer 55

ΔV is a change in gravitational potential, which is called a gravitational potential difference.

Question 56

What is required to move objects apart?

Answer 56

All masses attract each other. It takes energy, that is, external work must be done, to move objects apart. Gravitational potential is a maximum at infinity, where it’s value is taken to be 0Jkg^-1. This means that all values of gravitational potential are negative.

Question 57

What is gravitational potential, Vg at a point in a gravitational field?

Answer 57

The gravitational potential Vg at a point in a gravitational field is defined as the work done per unit mass to move an object to that point from infinity.

Question 58

How do you calculate the work done of an object being raised within a gravitational field

Answer 58

ΔW = mΔV

ΔV is a change in gravitational potential, which is called a gravitational potential difference

Question 59

What is GPE at infinity and how is work done related to it?

Answer 59

GPE is defined as being zero at infinity, and measured from there.

The gravitational potential energy (W) at a point in a field, is the work done to move an object from infinity to that point.

Question 60

Equation for Energy Change in a Magnetic Field

Answer 60

ΔE = m g Δh

The object has ΔE more potential for motion, due to being higher up in the gravitational field.

Question 61

How is gravitational potential, V, at a point in a gravitational field related to the amount of GPE

Answer 61

The gravitational potential, V, at a point in a gravitational field is the amount of GPE an object will have per unit mass

Question 62

Equation for Potential Energy where the object of a mass m, is in the gravitational field of mass M

(+ why is Ep always negative)

Answer 62

Ep = -GMm / r

Ep is always negative, because work must be done to move m towards infinity, against the attraction of the other mass.

Question 63

Why is Ep (gravitational potential energy) always negative

Answer 63

Ep is always negative, because work must be done to move m towards infinity, against the attraction of the other mass.

Question 64

Equation for gravitational potential, V, involving Ep (GPE) and m (mass)

Answer 64

V = Ep / m

Question 65

Calculate the minimum work required to lift an astronaut of mass 80kg from the Earth’s surface to the height of the ISS (300 km).

Earth’s radius, r = 6400km
Earth’s mass, M = 6.0x10^24kg
G = 6.672x10^-11 Nm^-2kg^-2

Answer 65

V = -GM/r

V at Earth surface: r = 6.4x10^6m
= -(6.67x10^-11) x (6.0x10^24) / 6.4x10^6
= -6.25x10^7 Jkg^-1

V at ISS: r= 6.7x10^6m
= -5.97x10^7 Jkg^-1

ΔV = (6.25-5.97)x10^7 Jkg^-1
ΔV = 2.8x10^6 Jkg^-1

ΔW = m x ΔV
80kg x 2.8x10^6 Jkg^-1
Work = 2.24x10^8 J = 220MJ

Question 66

What does the gravitational potential at any point in a radial field around a point mass depend on?

Answer 66

• The distance r from the point mass producing the gravitational field to that point.
• The mass M of the point mass

Question 67

What is gravitational potential, Vg proportional to and inversely proportional to?

Answer 67

Vg ∝ M
Gravitational potential is directly proportional to mass

Vg ∝ 1/r
Gravitational potential is inversely proportional to distance

Question 68

What does moving towards a point mass result in, in terms of gravitational potential

Answer 68

Moving towards a point mass results in a decrease in gravitational potential (e.g, from -40MJkg^-1 to -60MJkg^-1, giving ΔVg = -20MJkg^-1)

Question 69

What does moving away a point mass result in, in terms of gravitational potential

Answer 69

Moving away from a point mass (towards infinity) results in an increase in gravitational potential (e.g, from -30MJkg^-1 to -20MJkg^-1, giving ΔVg = +10MJkg^-1)

Question 70

How does gravitational potential change as you move away from the Earth’s surface

Answer 70

The gravitational potential rises are you move away from the surface of the Earth, before reaching a maximum value (note this value is not zero) and falling again as the distance to the Moon reduces.

Question 71

What is Potential Gradient (ΔV/Δr)

Answer 71

This is the change in potential per metre
at a point within a gravitational field.

potential gradient = ΔV / Δr

unit: J kg^-1 m^-1

Question 72

What are Gravitational Equipotentials?

Answer 72

These are surfaces that join up points of equal potential.

• No work is done by gravitational force when a mass is moved along an equipotential surface.
• Equipotentials are always perpendicular to field lines.
• Examples include: contour lines on maps, sea level, the floor of a room, the bench top surface.

Question 73

How do you find escape velocity (+ what is the escape velocity equation?)

Answer 73

In order for the projectile to have just enough energy to leave the gravitational field, the loss of kinetic energy must equal the gain in gravitational potential energy.

1/2mv^2 = GMm/r

v = √(2GM/r)

Question 74

Example Question:

a) What work would be needed to remove the astronaut completely from the Earth’s gravitational field?

b) If this work came from a conversion of initial kinetic energy (the astronaut is projected from the Earth’s surface), what would be the astronaut’s initial velocity?

Answer 74

Energy = -GMm / r
a)
E surface = (-6.67x10^-11 x 6.0x10^24 x 80) / 6400x10^3
E surface = -5.0 x 10^9

E gone = 0

b)
1/2mv^2 - 1/2mu^2 = -5.0x10^9
40x0 - 1/2 x 80u^2 = -5.0x10^9
40u^2 = 5.0x10^9
u = 11183ms^-1
u = 1.12x10^4