Gravity

Question 1

1 AU = 1.5x10^11

Answer 1

distance from the earth to the sun

given in the databook

Question 2

Definition of gravitational field strength

Answer 2

as the gravitational force acting on a unit mass

Question 3

F = GMm/r^2

Answer 3

F = Attractive Force (N)
G = Gravitational constant (6.67x10^-11) (m^3 kg-1 s-2)
M = mass 1 (kg)
m = mass 2 (kg)
r = distance between masses (m)

Masses M and m are regarded as point masses it is assumed that all their mass is concentrated at their centre

Question 4

Gravity obeys the inverse square law

Answer 4

The value of g varies inversely with the square of the distance. I.e. if the distance between two objects doubles the gravitational force between them quarters

Question 5

Field lines about two astronomical objects

Answer 5

Check jotter

Question 6

F = GMm/r^2 = mv^2/r = mrw^2 = mr(2π/T)^2

Answer 6

F = Attractive Force (N)
G = Gravitational constant (6.67x10^-11) (m^3 kg-1 s-2)
M = mass 1 (kg)
m = mass 2 or mass (kg)
v = linear velocity (ms-1)
r = radius or distance between two objects
w = angular velocity (rad s-1)
π = pi
T = period (s)

Question 7

gravitational force =

GMm/r^2

Answer 7

centripetal force

mw^2r

Question 8

Gravitational POTENTIAL

V = - GM/r

the minus sign is present to show that gravitational fields always act towards the attracting mass

Answer 8

V =  gravitational potential ( J kg-1)
G = 6.67x10^-11 (m^3 kg-1 s-2)
M = mass of object causing gravitational field (kg)
r = distance from mass (m)

Question 9

V = gravitational potential

Answer 9

J kg-1

Question 10

Ep = Gravitational potential ENERGY

Answer 10

J

Question 11

Ep = Vm = - GMm/r

Answer 11

Ep = Gravitational potential ENERGY (J)
V = Gravitational potential (J kg-1)
G = 6.67x10^-11 (m^3 kg-1 s-2)
M = mass causing gravitational field (kg)
m - mass being considered in gravitational field (kg)
r = distance from centre of mass M (m)

Question 12

BOTH gravitational potential energy and gravitational potential have the value of

Answer 12

zero at infinity

Question 13

Definition of the gravitational potential of a point in space

Answer 13

as the WORK DONE in moving UNIT MASS from infinity to that point

Question 14

Energy is required to move mass between

Answer 14

two points in a gravitational field is independent of the path taken

Question 15

Definition of escape velocity

Answer 15

Escape velocity is the minimum velocity required to allow a mass to escape a gravitational field to infinity, where the mass achieves zero kinetic energy and maximum (zero) potential energy.

Question 16

Derive escape velocity

Answer 16

Total energy on planet’s surface = total energy at infinity = 0
Ek + Ep = 0
1/2mv^2 + ( - GMm/r ) = 0 where M is the mass of the planet and r it’s radius
v^2 = 2GM/r
so v = √2GM/r

Question 17

vₑₛ꜀ = √2GM/r

Answer 17

vₑₛ꜀ = escape velocity (ms-1)
G = 6.67x10^-11 (m^3 kg-1 s-2)
M = Mass of the planet (kg)
r = radius of the planet (m)

Question 18

what is -1.70x10^9 J of gravitational potential mean?

Answer 18

-1.70x10^9 Joules transferred in moving unit mass from infinity to that point

Question 19

Solar radius

Answer 19

is a unit of distance used to express the size of stars in astronomy relative to the Sun.

It is given in the databook.

Question 20

Why is the sun not a black hole

Answer 20

The radius is greater than the Schwarzschild radius

Question 21

GMm/r^2 = mg

Answer 21

G = 6.67x10^-11
m/M = mass (kg)
r = distance (m)
g = gravitational field strength (Nm-1)

Question 22

acceleration reduces with

Answer 22

height

Question 23

when calculating the orbit of a object in space you will use

Answer 23

the bigger mass
i.e. if you were calculating the orbit of the moon you would use the mass of the earth as it is earths gravitational pull.