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Module 5 Physics > 5.4 > Flashcards

5.4 Flashcards

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1
Q

What is gravity?

A

Gravity is the universal attractive force which acts between all matter.

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2
Q

What is G?

A

The universal gravitational constant.

Approx. 6.67x10-11 m3 kg-1 s-2

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3
Q

What can field lines tell you about a field?

A

The direction of the field and the strength of the field depending on the density of the field lines.

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4
Q

What is 𝘨?

A

● 𝘨 is the force per unit area in a uniform gravitational field.
● In a radial field the magnitude of 𝘨 is the proportionality constant at that point between force and mass.
● I.e. 𝘨 = GM/r^2

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5
Q

What is Newton’s law of Gravitation?

A

Newton’s law of gravitation states that two point masses attract each other with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them.

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6
Q

What is Kepler’s first law?

A

Kepler’s first law states that the orbit of a planet is an ellipse, with the sun at one focus. The eccentricity of the ellipse is very low, so the motion can be modelled as circular.

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7
Q

What is Kepler’s second law?

A

Kepler’s second law states that a line segment joining a planet and the sun sweeps out equal areas during intervals of equal time. This is because the speed of the planet is not constant – the planet moves faster when it is closer to the sun.

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8
Q

What is Kepler’s third law?

A

Kepler’s third law states that the square of the orbital period T is proportional to the cube of the average distance r from the sun. This can be proved by considering the forces acting on the planet. Centripetal force is required to keep the planet in orbit, and this force is provided by the gravitational field of the sun.

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9
Q

Derive this equation to show T2 is proportional to r3 and explain your steps
T^2 = 4π^2r^3/GM

A
  1. Because of Kepler’s third law, we can equate the formula for centripetal force with the formula for gravitational force to get mv^2/r = GMm/r^2
  2. Rearrange to get v^2 = GM/r
  3. Since velocity in circular motion is 2πr/T, you can substitute this into the previous equation to get 4π^2r^2/T^2 = GM/r
  4. Rearrange this to get T^2 = 4π^2r^3 / GM
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10
Q

What are satellites? What are they used for?

A

● Satellites are objects that orbit other, larger objects. These can include natural satellites like the moon, and artificial satellites that humans have sent into space.
● Uses include: communications, scientific
research, and Global Positioning Systems (GPS).

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11
Q

What are geostationary satellites? What are they used for?

A

● Geostationary satellites have an orbital period that is exactly a day, so that they appear stationary above the Earth.
● They orbit 36,000km above the equator.
● They are useful for communications and surveying as they provide continuous coverage.

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12
Q

What is gravitational potential?

A

The potential energy per kilogram, at any point in the field.
0 potential is defined at infinity, so at a point close to a mass, the potential of an object would be negative.

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13
Q

What is gravitational potential difference?

A

Gravitational potential difference is the difference in the gravitational potentials of two points in a gravitational field.

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14
Q

What is gravitational potential energy at a point in the field?

A

The work done per unit mass in moving object with from infinity to that point in the field.

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15
Q

What is escape velocity?

A

● The minimum velocity an object requires in order to
escape the gravitational field of an object when
projected vertically from its surface.
● The formula for vesc is derived from equating the
kinetic energy and the gravitational potential energy

required to reach infinity: ½ mv^2 = GMm/r
○ Rearrange this to get v = sqrt(2GM/r)

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16
Q

Explain what causes a body to have gravitational fields.

A

The property of mass causes a body to have gravitational fields.

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17
Q

Explain why the force of gravitational attraction is not noticeable between two cars but it is noticeable between two planets.

A

The mass of two cars is far smaller than the mass of two planets. Therefore, as the gravitational force is directly proportional to the product of the masses, the gravitational force between the lighter cars is much smaller than that between the two planets.

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18
Q

What do gravitational radial field lines tell us about the nature of a graviational field?

A

The gravitational force is attractive.
The gravitational field is radial.
The gravitational field acts towards the centre of Mars.
The gravitational field decreases with distance from the surface of Mars.

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19
Q

How does the strength of the gravitational field vary with distance from the surface of Earth?
Explain what feature of a radial diagram shows this.

A

The gravitational field strength decreases with distance.
We can see this on the diagram as the field lines are closer nearer to the surface and become farther apart as the distance increases.
The closer together the field lines, the stronger the gravitational field.

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20
Q

Write and equation for gravitational field strength F and m.

A

g = F/m

g - gravitational field strength

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21
Q

Define the gravitational field strength.

A

The gravitational field strength is defined as the force per unit mass experienced by a small mass at a point within the field

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22
Q

Name two fields that can give rise to a force (excluding gravitational fields)

A

Electric fields and magnetic fields.

23
Q

Explain the importance of the point mass

A

Spheres can be moddelled as a point mass at their centre. Due to the sphere’s symmetry we are able to simplify and model it as a point mass, simplifying the mathematics.

24
Q

Write and equation for gravitational field strength involving G.

A

g = - GM/r^2

g - gravitational field strength at a point
G- gravitational constant (6.67*10-11)
M - mass of body causing the field
r - separation between point and the centre of the body causing the gravitational field

25
Q

What is the graph that shows how the magnitude of the gravitational force varies with distance from the centre of a sphereical body of radius R.

A

Gravitational force against distance from centre of earth, inversely proportional, asymptote at x = R

26
Q

Derive g = - GM/r^2

A

We know that:
F = -GMm/r^2

and that:
F = mg

Equating the two:
mg = GMm/r^2

Cancelling m:
g = -GM/r^2

27
Q

We know that g obeys an inverse square relationship with distance.

Explain why there is no noticeable difference in the value of g at the heights reached by aeroplanes.

A

The distance is measured from the centre of the body.

Heights from the surface of the Earth reached by aeroplanes are very small in comparison to the distance from the centre of the Earth to the surface.

Therefore, this increase in distance from the centre of the Earth only has a small effect on the value of g.

28
Q

State a unit that is equal to N/kg.

Show how this is true.

A

The unit is:
ms^-2

The two units are equivalent because:
F=ma

We know the g is a form of acceleration.
a = g

Therefore, by rearranging we get:
g = F/m -> N/kg

We also know that a = (v-u)/t -> ms^-2

therefore the two units are equivalent.

29
Q

State Newton’s law of gravitation.

A

The force of attraction between two particles is directly proportional to the product of the masses of the two particles and inversely proportional to the square of the distance between the centres of the two objects.

F=-(GMm)/r^2

30
Q

Describe the gravitational field and the gravitational field strength close to the surface of the Earth.

A

The gravitational field strength is approximate4 equal to 9.81M/kg close to the surface of the Earth and the field lines are approximately parallel.

31
Q

Explain how the gravitational force obeys the inverse square law.

A

As the distance from the centre of the Earth doubles the gravitational force gets four times smaller.

32
Q

State Kepler’s first law.

A

The orbit of every planet is an ellipse with the Sun at the foci.

33
Q

State Kepler’s second law.

A

The line between the centre of the Sun and the centre of a planet will sweep out equal areas in equal times.

34
Q

State Kepler’s third law

A

The square of a planet’s period is directly proportional to the cube of its average distance from the Sun.

35
Q

Explain why a planet orbitting the Sun in a circular orbit can have a constant speed but not have a constant velocity.

A

As the orbit is circular, the planet’s motion is constantly changing direction. Velocity is a vector so it has magnitude and direction; however, speed is simply a scalar.

36
Q

Explain why a planet orbitting the Sun in a circular orbit can have a constant speed but not have a constant velocity.

A

As the orbit is circular, the planet’s motion is constantly changing direction. Velocity is a vector so it has magnitude and direction; however, speed is simply a scalar.

The magnitude of speed and velocity can remain the same but because its direction is changing it has constant speed but not velocity.

37
Q

State the direction of the Earth’s acceleration as it orbits the Sun.

A

Directed towards the Sun.

38
Q

A planet is orbiting the Sun. Describe the forces acting on it.

A

The only force acting on the planet is gravitational attraction.

39
Q

Explain why planets move faster at smaller orbits.

A

According to Kepler’s third law, the square of a planet’s period is directly proportional to the cube of its average distance from the Sun.

This means that if the planet has a smaller orbit as it is nearer to the Sun, it will also have a smaller period.

We know that:
v=2(pi)/T

Therefore, the smaller the period the bigger the velocity.

40
Q

What does a graph of T^2 against r^3.

A

Directly proportional

41
Q

Give three uses of geostationary satellites.

A

Spying
Telecommunication
Weather

42
Q

State the properties of a geostationary orbit.

A

The orbit is above the equator.
It stays at the same position relative to the surface of the Earth.
It has an orbital period equal to 24 hrs.

43
Q

Define the gravitational potential at a point.

A

The work done in bringing a unit mass from infinity to the point.

44
Q

Explain why gravitational potential is relative to zero.

A

At all points at a distance of infinity the potential energy is equal to zero.

45
Q

Write down an equation for gravitational potential.

A

Vg = - GM/r

46
Q

State three factors that potential energy is dependent on.

A
  1. The mass of object 1
  2. The mass of object 2
  3. The separation of the two objects
47
Q

Explain how the gravitational force and its relationship with gravitational field strength is similar to that between potential and potential energy.

A

They both share a similar relationship in that
F = mg

and
E = mV(g)

to find the gravitational force/ potential energy we multiply the gravitational field strength/ potentail by the mass.

48
Q

Force - distance graph for a point mass

A

inversely proportional, flipped in x axis, asymptote at x=0

49
Q

Force - distance graph for a spherical mass of radius R

A

inversely proportional, with asymptote at x=R, reflected in x axis. Straight line drawn from peak of asymptote at R to origin,

50
Q

What are the two equations for gravitational potential energy

A

E = mV(g)

E = -GMm/r

r  = separation
V(g) = gravitational potential
51
Q

Unit of gravitational potential is J/kg. Derive this unit.

A

E=mV(g)

therefore V(g) = E/m

which equals Jkg^-1

52
Q

Define escape velocity

A

The escape velocity is the minimum velocity an object must have to be able to overcome the gravitational attraction acting on it.

53
Q

What is the equation for escape velocity.

A

v = sqrt(2GM/r)

54
Q

Derive the equation for the escape velocity

A

E = GMm/r

E = 1/2mv^2

1/2mv^2 = GMm/r

1/2v^2 = GM/r

v = sqrt(2GM/r)

Module 5 Physics (4 decks)

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