Module 5 Gravitational Fields

Question 1

Define a gravitational field and gravitational field lines.

Answer 1

A region around a mass where other masses feel an attractive force towards the mass, is a gravitational field.

Gravitational field lines show the direction of the force of gravity.

Question 2

Can there be a point between two bodies of large mass where no net force occurs?
Why?

Answer 2

Yup.

The gravitational fields of the two masses cancel each other out at specific points.

Question 3

What quantity causes gravity?

Answer 3

Mass.

Question 4

For a spherical mass, where do the gravitational field lines point?

Answer 4

Towards the centre of the circle.

Question 5

What is the distance limit of gravitational force?

Answer 5

It has no limits, as gravitational force has an infinite range.

Question 6

What is the relation between gravitational field strength and separation.
Give an example.

Answer 6

G.f.s is inversely proportional to the square of separation.

If d is increased by 3 times then the g.f.s is decreased by 9 times.

Question 7

Why can gravitational field lines be considered parallel on the surface on Earth?

Answer 7

As the variant in g.f.s is minuscule, we assume that g.f.s is uniform over the surface.

Question 8

Is gravitational field a vector?

Answer 8

Yup

Question 9

Define gravitational field strength and state why it is a vector quantity?
Give equations and units too.

Answer 9

G.f.s is the force acting on a body divided by it’s mass at any point in gravitational field.

It is a vector since it is the product of a vector and a scalar(F x 1/m) hence it is a vector quantity.

g=F/m
g=Nkg^-1

Question 10

Even though parallel lines can be drawn on the surface of the Earth to show g.f lines, in reality are there any fluctuations in g.f.s and why?

Answer 10

Yes there are tiny fluctuations as the matter across the surface of Earth is not uniform. The composition of Earth varies at different points and this affects the mass of Earth at that point, causing difference in g.fields.

Question 11

What is the difference and similarities between acceleration g= 9.81 and gfs on the surface of the Earth.

Answer 11

“g” is the acceleration due to gravity and has units ms^-2, whereas gfs is has the units N/kg.

They both have equal magnitudes.

Question 12

When you are stating that a 10 kg mass has a weight of 98.1N, are you multiplying mass with g or g.f.s?

Answer 12

G.f.s

Question 13

State Newton’s law of gravitation and give equations.

Answer 13

Between two masses the force of attraction is directly proportional to the product of their masses and inversely proportional to the square of their separation.

F=-GMm/r^2 , where G is universal gravitational constant, M and m are masses of respective bodies and r is the separation between the CENTRE of the bodies

Question 14

Why is there a negative sign in the equation for Newton’s law of gravitation.

Answer 14

This is to show that gravity is always attractive.

Question 15

Relate the g.f.s of a point mass to Newton’s law of gravitation.

Answer 15

Since g = F/m

F=-GMm/r^2

g=-GMm/r^2 m
g=-GM/r^2

Question 16

Describe the curve of an inverse square relationship between g and r.

Answer 16

If y axis is the quantity inversely proportional to square of another value, then the graph has increasing gradient but the increase in gradient becomes smaller.
(Makes the graph look like a boomerang with the outward tip pointed towards the origin)

Question 17

Define gravitational potential. What is it’s sign?

Answer 17

At a point in a gravitational field, it is the work done to move a unit mass from infinity to that specific point.
Vg.

Question 18

State equation and units for Vg.

Answer 18

Vg=-GM/r

Vg=Jkg^-1

Question 19

How do you calculate change in Vg, when separation is changed?

Answer 19

Calculate initial vg and final vg and find the difference.

Question 20

What is the magnitude of Vg at infinity?

How do you calculate change in Vg?

Answer 20

It is zero 0.

You first calculate the Vg at each point then find the difference between them.

Question 21

Define the gradient of g.f.s against separation, for a distance from the centre of the sphere to the surface.

Answer 21

It is a straight line graph through the origin, meaning r and g are directly proportional within a spherical mass.

Question 22

What is the gradient of a Vg against r graph equal to?

Answer 22

It is equal to -g

Question 23

What is the area under a F against r graph?

Answer 23

It is equal to the work done in moving a body in a gravitational field.

Question 24

Define gravitational potential energy.

Derive equations.

Answer 24

Gravitational potential energy for a defined mass ‘m’ of a body is the product of it’s mass and the gravitational potential.

E=m x Vg

Vg=-GM/r

E=-GMm/r

Question 25

Derive equation to calculate the speed of an orbiting body.

Answer 25

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

-GMm/r^2= -mv^2/r
-GMm/r=-mv^2
v^2=-GMm/mxr
v^2=-GM/r
v=(-GM/r)^1/2

Question 26

What is the change in Gravitational potential energy for a body travelling at escape velocity from the surface of the Earth?

Answer 26

It is increasing from E= -GMm/r to 0

Question 27

Define escape velocity and Derive equation for escape velocity.

Answer 27

It is the velocity required to move a body from a point in a grav. field to infinity.

For a body travelling at escape v
K.E>= Gravitational potential energy

1/2xm v^2 = -GMm/r
v^2= -2 GMm/m r
v= (-2 x GM/r)^1/2

Question 28

State Kepler’s 3 laws of planetary motion.

Answer 28

The planets have an elliptical orbit around the Sun.

A line joining the Sun and any planet will sweep equal amounts of area for the same time.

The square of the Time taken to orbit the Sun is directly proportional to the mean radius of the orbit cubed, for any planet in the Sun’s orbit.

Question 29

Derive the equation for Time period for a planet.

Answer 29

v=d/t
t=T
T=d/v

T=2pi r/v

v=(GM/r)^1/2 (v for an orbiting body)

T=2 pi r/ (GM/r)^1/2
T^2= 4 pi^2 r^2/ GM/r
T^2= 4 pi^2 r^3/GM
T= (4 pi^2 r^3/GM)

Question 30

Derive T for a body in orbit using v =2pi r/T

Answer 30

v=2 pi r/T, F=mv^2/r

F=m 4 pi^2 r^2/T^2 x r = m 4 pi^2 r/T^2

F=-GMm/r^2

m 4pi^2 r/T^2= -GMm/r^2
T^2=m 4 pi^2 r/ -GMm/r^2
T^2=-4 pi^2 r^3/GM

Question 31

What causes the centripetal force in orbits?

Answer 31

The gravitational force as it is perpendicular to the direction of motion but acts towards the centre of the larger mass

Question 32

Define a geostationary orbit.

Answer 32

An orbit that has the same rotational time period as the body it’s orbiting, 24 hrs for Earth is a geostationary orbit. The direction of orbit is also the same as the direction of rotation.