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Physics A Level CIE > Chapter 2 Gravitational Fields > Flashcards

Chapter 2 Gravitational Fields Flashcards

1
Q

Defining Gravitational Field

A
  • 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
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2
Q
  • A gravitational field is defined as:
A
  • 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
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3
Q

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

A
  • Where: F=ma
  • g = gravitational field strength (N kg-1)
  • Fg = force due to gravity, or weight (N)
  • m = mass (kg)
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4
Q

Representing Gravitational Fields

A
  • 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
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5
Q

Radial fields are considered

A
  • are considered non-uniform fields
    • The gravitational field strength g is different depending on how far you are from the centre
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6
Q
  • Parallel field lines on the Earth’s surface are considered a
A
  • uniform field
    • The gravitational field strength g is the same throughout
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7
Q

Point Mass Approximation

A
  • 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
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8
Q
  • An object can be regarded as point mass when:
A
  • 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
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9
Q

Newton’s Law of Gravitation

A
  • 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:
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10
Q
  • Newton’s Law of Gravitation states that:
A

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

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11
Q

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

A

Gravitation Equation

  • Where:
    • FG = 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)
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12
Q

The inverse relationship

A
  • 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
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13
Q

Circular Orbits in Gravitational Fields

A
  • Since most planets and satellites have a near circular orbit, the gravitational force FG 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
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14
Q

Kepler’s Third Law of Planetary Motion

A
  • 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
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15
Q
  • 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:
A

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:
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16
Q

Circular Orbits in Gravitational Fields equations

A
  • Consider a satellite with mass m orbiting Earth with mass M at a distance r from the centre travelling with linear speed v
  • Fg=Fcirc
  • 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
17
Q

Geostationary Orbits

A
  • 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
18
Q

use of geostationary orbits

A
  • 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
19
Q

Deriving Gravitational Field Strength (g)

A
  • 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:

Fg=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 FG 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)
20
Q

Newton’s Gravitational Constant

A
  • 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

21
Q

Calculating g

A
  • 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)
22
Q

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

A
23
Q

The Value of g on Earth

A
  • 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
24
Q

Gravitational Potential

A
  • 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
25
Q
  • However, outside the Earth’s surface, G.P.E can be defined as:
A

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

26
Q

Calculating Gravitational Potential

A
  • 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)
27
Q

Point Mass

A

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

28
Q

The gravitational potential is negative near

A
  • 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
29
Q

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

A
30
Q

Gravitational Potential Energy Between Two Point Masses

A
  • 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)
31
Q

Gravitational Potential

A

The work done per unit mass in bringing a test mass from infinity to a defined point.

  • Φ = -GM/r
32
Q

Recall that at infinity, ɸ =

A
  • 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

33
Q

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

A

Physics A Level CIE (14 decks)

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