Gravity Flashcards
1 AU = 1.5x10^11
distance from the earth to the sun
given in the databook
Definition of gravitational field strength
as the gravitational force acting on a unit mass
F = GMm/r^2
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
Gravity obeys the inverse square law
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
Field lines about two astronomical objects
Check jotter
F = GMm/r^2 = mv^2/r = mrw^2 = mr(2π/T)^2
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)
gravitational force =
GMm/r^2
centripetal force
mw^2r
Gravitational POTENTIAL
V = - GM/r
the minus sign is present to show that gravitational fields always act towards the attracting mass
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)
V = gravitational potential
J kg-1
Ep = Gravitational potential ENERGY
J
Ep = Vm = - GMm/r
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)
BOTH gravitational potential energy and gravitational potential have the value of
zero at infinity
Definition of the gravitational potential of a point in space
as the WORK DONE in moving UNIT MASS from infinity to that point
Energy is required to move mass between
two points in a gravitational field is independent of the path taken
Definition of escape velocity
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.
Derive escape velocity
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
vₑₛ꜀ = √2GM/r
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)
what is -1.70x10^9 J of gravitational potential mean?
-1.70x10^9 Joules transferred in moving unit mass from infinity to that point
Solar radius
is a unit of distance used to express the size of stars in astronomy relative to the Sun.
It is given in the databook.
Why is the sun not a black hole
The radius is greater than the Schwarzschild radius
GMm/r^2 = mg
G = 6.67x10^-11 m/M = mass (kg) r = distance (m) g = gravitational field strength (Nm-1)
acceleration reduces with
height
when calculating the orbit of a object in space you will use
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.
