5 Gravitational fields

Question 1

gravitational fields

Answer 1

due to objects having mass
all objects with mass create a gf around them which extends all the way to infinity, but gets weaker as the distance from the centre of mass of the object increases
any other object with mass placed in a gf will experience an attractive force towards the centre of mass of the object decreasing the field

Question 2

on earth what is an object’s gravitational attraction called?

Answer 2

weight

Question 3

radial field of a planet diagram

Answer 3

when the gf pattern is drawn, it has straight lines converging to the centre of mass.
The lines are arrows pointing to the centre of mass, to show that gravity is an attractive force.
The closer the lines are to each other, the stronger the field is at that point. The lines never cross each other.

Question 4

uniform gf diagram

Answer 4

such as looking at the surface of a planet on a small scale.
Here, the field strength is equal at all positions, so the field is drawn as parallel lines towards the surface, at equal intervals from each other.

Question 5

gravitational field strength

Answer 5

the gravitational force experienced per unit mass by an object at that point in a gravitational field.
It is a vector quantity, with the units Nkg-1 or ms-2.
g=F/m
F is gf
m is mass of object
This equation is accurate as long as the mass of the object in the field is small enough that the
object’s gravitational field is negligible compared to the external gravitational field the object
is in.

Question 6

gfs equation

Answer 6

g=F/m
F is gf
m is mass of object
This equation is accurate as long as the mass of the object in the field is small enough that the object’s gravitational field is negligible compared to the external gravitational field the object is in.

Question 7

point mass

Answer 7

All objects with mass can be modelled as a point mass, where the point is the centre of mass of the object.

Question 8

newton’s law of gravitation

Answer 8

states 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 their separation. This can be written to show that the gravitational force, F, between two objects is
F= -GMm/r2

Where G is the gravitational constant, 6.67 × 10−11 Nm2kg-2, M and m are the masses of the
two objects, and r is the distance between their centres. The negative sign is used to show the attractive nature of the force.

Question 9

inverse-square relationship between force and distance

Answer 9

the attractive force F between objects decreases with distance in an inverse-square relationship
F dir prop 1/r2

Question 10

gfs for a point mass

Answer 10

As the gravitational field strength is given by 𝑔=𝐹/𝑚, the gfs for a point mass can be given by dividing the gravitational force between two point masses by the mass of the other point mass. This produces
g=-GM/r2

This shows that the field strength for an object, such as a planet, does not depend on the mass of the object in orbit around the planet, only on the mass of the planet and the distance between them.
When considering the gravitational field strength close to the earth’s surface, the field can be modelled as uniform, and has the same value as the acceleration of free fall.

Question 11

relationship between gfs and mass of object creating the gf

Answer 11

g dir prop M

Question 12

relationship between gfs and distance from centre of mass of the object

Answer 12

g dir prop 1/r2

inversely prop to the square of the distance

Question 13

graph of gfs against distance from the centre of mass of object creating a gfs

Answer 13

values of g are negative
as the distance from the centre of mass increases, g decreases until, at infinity, it reaches zero
graph of g against 1/r2 is a straight line through the origin with a gradient equal to -GM

Question 14

kepler’s first law

Answer 14

the orbit of a planet is an ellipse with the sun at one of two foci

The eccentricity (measure of how elongated the circle is) of the ellipse is very low, so the motion can be modelled as circular.

Question 15

kepler’s second law

Answer 15

a line segment joining a planet and sun sweeps out equal areas during equal intervals of time

This is because the speed of the planet is not constant – the planet moves faster when it is closer to the sun.

Question 16

kepler’s third law

Answer 16

the square of the orbital period T is directly proportional to the cube of the average distance r from the sun. T
T2/r3 = k

Question 17

centripetal force

Answer 17

=gravitational force on planet

Question 18

putting a satellite into orbit

Answer 18

only force acting on a satellite is the gravitational attraction between it and and the earth, it’s always falling towards the earth
however as it is travelling so fast, it travels such a great distance that as it falls the earth curves away beneath it, keeping it at the same height above the surface
all satellites must therefore be given exactly the right height and speed for a stable orbit
v= square root GM/r

Question 19

use of satellites

Answer 19

• communications: satellite phones, TV
• military uses : reconnaissance
• scientific research
• weather and climate
• global positioning

Question 20

types of orbit

Answer 20

polar orbits- circles the poles
equatorial orbits
low earth orbits

Question 21

geostationary satellites

Answer 21

placed in a specific orbit so that it remains above the same point of the earth whilst the earth rotates
for this, the satellite must:
-be in an orbit above the earth’s equator
-rotate in the same direction as the earth’s rotation
-have an orbital period of 24hrs

Question 22

gravitational potential

Answer 22

the gp (Vg) at a point in a gf is defined as the work done per unit mass to move an object to that point from infinity
unit Jkg-1
scalar quantity

all masses attract each other. it takes energy, that is, external work must be done, to move objects apart.
gp is a maximum at infinity, where its value is taken to be 0Jkg-1
so all values of gp are negative

Question 23

infinity

Answer 23

refers to a distance so far from the object producing the gf that the gfs is zero

Question 24

gp at any point in a radial field around a point mass depends on what two factors?

Answer 24

• distance r from the point mass producing the gf to that point
• mass M of the point mass

Vg dir prop M
Vg die prop 1/r

Question 25

graph of Vg against r

Answer 25

Vg dir prop 1/r
the potential will tend towards zero as r approaches infinity
the smallest value of r must be equal to the radius of the earth

graph of Vg against 1/r produces a straight line through the origin with gradient equal to -GM

Question 26

changes in gp

Answer 26

moving from one point in a gf to another results in a change in gp

• moving towards a point mass results in decrease in gp
• moving away from point mass results in increase in gp

Question 27

gravitational potential energy

Answer 27

defined as work done to move the mass from infinity to a point in a gf
E= m Vg

Question 28

gpe in a uniform gf

Answer 28

in order to change the gpe of an object, its height above the surface must be changed
this results in a change in gp and so a change in gpe

Question 29

escape velocity

Answer 29

in order to escape the gf of a mass like a planet, an object must be supplied with energy equal to the gain in gpe needed to lift it out of the field

1/2mv2=GMm/r
v= square root 2GM/r

the escape velocity on a given planet is therefore the same for all objects regardless of their mass