Chapter 2 Gravitational Fields

Question 1

Defining Gravitational Field

Answer 1

• There is a force of attraction between all masses
• This force is known as the ‘force due to gravity’ or the weight
• The Earth’s gravitational field is responsible for the weight of all objects on Earth

Question 2

• A gravitational field is defined as:

Answer 2

• A region of space where a mass experiences a force due to the gravitational attraction of another mass
• The direction of the gravitational field is always towards the centre of the mass
• Gravitational forces cannot be repulsive

Question 3

The strength of this gravitational field (g) at a point is the force (F_g) per unit mass (m) of an object at that point:

Answer 3

• Where: F=ma
• g = gravitational field strength (N kg^-1)
• F_g = force due to gravity, or weight (N)
• m = mass (kg)

Question 4

Representing Gravitational Fields

Answer 4

• The direction of a gravitational field is represented by gravitational field lines
• The gravitational field lines around a point mass are radially inwards
• The gravitational field lines of a uniform field, where the field strength is the same at all points, is represented by equally spaced parallel lines
  
  • For example, the fields lines on the Earth’s surface

Question 5

Radial fields are considered

Answer 5

• are considered non-uniform fields
  
  • The gravitational field strength g is different depending on how far you are from the centre

Question 6

• Parallel field lines on the Earth’s surface are considered a

Answer 6

• uniform field
  
  • The gravitational field strength g is the same throughout

Question 7

Point Mass Approximation

Answer 7

• For a point outside a uniform sphere, the mass of the sphere may be considered to be a point mass at its centre
  
  • A uniform sphere is one where its mass is distributed evenly
• The gravitational field lines around a uniform sphere are therefore identical to those around a point mass

Question 8

• An object can be regarded as point mass when:

Answer 8

• A body covers a very large distance as compared to its size, so, to study its motion, its size or dimensions can be neglected
• So, the gravitational field strength (The force per unit mass on an object in a gravitational field. g = 9.81 N/kg on Earth. g = GM/r²
• g is different depending on how far you are from the centre of mass of the sphere

Question 9

Newton’s Law of Gravitation

Answer 9

• The gravitational force between two bodies outside a uniform field, e.g. between the Earth and the Sun, is defined by Newton’s Law of Gravitation:

Question 10

• Newton’s Law of Gravitation states that:

Answer 10

The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square their separation

Question 11

Newton’s Law of Gravitation equation form, this can be written as:

Answer 11

Gravitation Equation

• Where:
  
  • F_G = gravitational force between two masses (N)
  • G = Newton’s gravitational constant
  • m₁ and m₂ = two points masses (kg)
  • r = distance between the centre of the two masses (m)

Question 12

The inverse relationship

Answer 12

• Although planets are not point masses, their separation is much larger than their radius
  
  • Therefore, Newton’s law of gravitation applies to planets orbiting the Sun
• 1/r²
• This means that when a mass is twice as far away from another, its force due to gravity reduces by (½)² = ¼

Question 13

Circular Orbits in Gravitational Fields

Answer 13

• Since most planets and satellites have a near circular orbit, the gravitational force F_G between the sun or another planet provides the centripetal force needed to stay in an orbit
• Both the gravitational force and centripetal force are perpendicular to the direction of travel of the planet

Question 14

Kepler’s Third Law of Planetary Motion

Answer 14

• For the orbital time period T to travel the circumference of the orbit 2πr, the linear speed v can be written as

v=2πr/T

• This is a result of the well-known equation, speed = distance / time
• Substituting the value of the linear speed v into the above equation:
• v²= (2πr/T)²=GM/r
• Rearranging leads to Kepler’s third law equation:
• T²=4(π)²r³/GM

Question 15

• The equation shows that the orbital period T is related to the radius r of the orbit. This is known as Kepler’s third law:

Answer 15

For planets or satellites in a circular orbit about the same central body, the square of the time period is proportional to the cube of the radius of the orbit

• Kepler’s third law can be summarised as:

Question 16

Circular Orbits in Gravitational Fields equations

Answer 16

• Consider a satellite with mass m orbiting Earth with mass M at a distance r from the centre travelling with linear speed v
• F_g=F_circ
• Equating the gravitational force to the centripetal force for a planet or satellite in orbit gives:
• GMm/r² = mv²/r
• The mass of the satellite m will cancel out on both sides to give:
• v²=GM/r
• This means that all satellites, whatever their mass, will travel at the same speed v in a particular orbit radius r
• Recall that since the direction of a planet orbiting in circular motion is constantly changing, it has centripetal acceleration

Question 17

Geostationary Orbits

Answer 17

• Many communication satellites around Earth follow a geostationary orbit
• This is a specific type of orbit in which the satellite:
  
  • Remains directly above the equator, therefore, it always orbits at the same point above the Earth’s surface
  • Moves from west to east
  • Has an orbital time period equal to Earth’s rotational period of 24 hours

Question 18

use of geostationary orbits

Answer 18

• Geostationary satellites are used for telecommunication transmissions (e.g. radio) and television broadcast
• A base station on Earth sends the TV signal up to the satellite where it is amplified and broadcast back to the ground to the desired locations
• The satellite receiver dishes on the surface must point towards the same point in the sky
  
  • Since the geostationary orbits of the satellites are fixed, the receiver dishes can be fixed too

Question 19

Deriving Gravitational Field Strength (g)

Answer 19

• For calculations involving gravitational forces, a spherical mass can be treated as a point mass at the centre of the sphere
• Newton’s law of gravitation states that the attractive force F between two masses M and m with separation r is equal to:

F_g=GMm/r²

• The gravitational field strength at a point is defined as the force F per unit mass m

g=F/m

• Substituting the force F with the gravitational force F_G leads to:

g=F/m=((GMm)/r²)÷m

• Cancelling mass m, the equation becomes:

g=GM/r²

Where:

• g = gravitational field strength (N kg^-1)
• G = Newton’s Gravitational Constant
• M = mass of the body producing the gravitational field (kg)
• r = distance from the mass where you are calculating the field strength (m)

Question 20

Newton’s Gravitational Constant

Answer 20

• a constant used to relate the gravitational field strength between two objets to their mass and separation. the number will be given to you in the exam.

G= 6.67 x 10^-11

Question 21

Calculating g

Answer 21

• Gravitational field strength, g, is a vector quantity
• The direction of g is always towards the centre of the body creating the gravitational field
  
  • This is the same direction as the gravitational field lines
• On the Earth’s surface, g has a constant value of 9.81 N kg^-1
• However outside the Earth’s surface, g is not constant
  
  • g decreases as r increases by a factor of 1/r²
  • This is an inverse square law relationship with distance
• When g is plotted against the distance from the centre of a planet, r has two parts:
  
  • When r < R, the radius of the planet, g is directly proportional to r
  • When r > R, g is inversely proportional to r² (this is an ‘L’ shaped curve and shows that g decreases rapidly with increasing distance r)

Question 22

Graph showing how gravitational field strength varies at greater distance from the Earth’s surface

Answer 22

Question 23

The Value of g on Earth

Answer 23

• Gravitational field strength g is approximately constant for small changes in height near the Earth’s surface (9.81 m s^-2)
• This is because from the inverse square law relationship:

g∝1/r²

• The value of g depends on the distance from the centre of Earth r
• If we take a position h above the Earth’s surface, where it is reasonable to assume h is much smaller than the radius of the Earth (h << R):

g=GM/(R+h)²≈GM/(R)²

• This means g remains approximately constant until a significant distance away from the Earth’s surface

Question 24

Gravitational Potential

Answer 24

• The gravitational potential energy (G.P.E) is the energy an object has when lifted off the ground given by the familiar equation:
• G.P.E = mgΔh
• The G.P.E on the surface of the Earth is taken to be 0
  
  • This means work is done to lift the object

Question 25

* However, outside the Earth’s surface, G.P.E can be defined as:

Answer 25

***The energy an object possess due to its position in a gravitational field*** * The gravitational potential at a point is the gravitational potential energy per unit mass at that point * Therefore, the gravitational potential is defined as: ***The work done per unit mass in bringing a test mass from infinity to a defined point***

Question 26

**Calculating Gravitational Potential**

Answer 26

* The equation for gravitational potential ɸ is defined by the mass *M* and distance *r*: ![]() * Where: * ɸ = gravitational potential (J kg^-1) * G = **Newton’s gravitational constant** * M = mass of the body producing the gravitational field (kg) * r = distance from the centre of the mass to the **point mass**(m)

Question 27

Point Mass

Answer 27

When a body covers a very large distance as compared to its size, its size or dimensions can be neglected to study its motion.

Question 28

The gravitational potential is negative near

Answer 28

* an isolated mass, such as a planet, because the **potential when** ***r*** **is at infinity is defined as 0** * Gravitational forces are always **attractive** so as *r* decreases, positive work is done by the mass when moving from infinity to that point * When a mass is closer to a planet, its gravitational potential becomes smaller (more negative) * As a mass moves away from a planet, its gravitational potential becomes larger (less negative) until it reaches 0 at infinity * This means when the distance (*r)* becomes very large, the gravitational force tends rapidly towards 0 at a point further away from a planet

Question 29

***Gravitational potential increases and decreases depending on whether the object is travelling towards or against the field lines from infinity***

Answer 29

Question 30

**Gravitational Potential Energy Between Two Point Masses**

Answer 30

* The gravitational potential energy (G.P.E) at point in a gravitational field is defined as: ***The work done in bringing a mass from infinity to that point*** * The equation for G.P.E of two point masses *m* and M at a distance *r* is: * **GPE=-GMm/r** * The change in G.P.E is given by: * **ΔG.P.E = mgΔh** * Where: * m = mass of the object (kg) * ɸ = **gravitational potential** at that point (J kg^-1) * Δh = change in height (m)

Question 31

Gravitational Potential

Answer 31

The **work done per unit mass** in bringing a test mass from infinity to a defined point. * **Φ = -GM/r**

Question 32

Recall that at infinity, ɸ =

Answer 32

* 0 and therefore G.P.E = 0 * It is more useful to find the change in G.P.E e.g. a satellite lifted into space from the Earth’s surface * The change in G.P.E from for an object of mass *m* at a distance r₁ from the centre of mass *M*, to a distance of r₂ further away is: **ΔGPE= -GMm/r₂-(-GMm/r₁)=GMm(1/r₁-1/r₂)** ***Change in gravitational potential energy between two points*** * The change in potential Δɸ is the same, without the mass of the object *m*: **Δɸ= -GM/r₂-(-GM/r₁)=GM(1/r₁-1/r₂)** ***Change in gravitational potential between two points***

Question 33

***Gravitational potential energy increases as a satellite leaves the surface of the Moon***

Answer 33