M5, C3 Gravitational Fields Flashcards
define gravitational field
a force field generated by any object with mass which causes any other object with mass to experience an attractive force
what are gravitational field lines
how would you draw them
known as lines of force
show the direction of the force that masses would feel in a gravitational field
drawn towards the edge of the object with arrows pointing inwards
what is Newton’s law of gravitation
the force acting between two point masses is proportional to the product of their masses and inversely proportional to the square of the distance between their centres of mass
F = - GMm / r^2
what does this equation mean
F = force acting on mass m due to mass M G = gravitational constant (= 6.67 X 10^-11 Nm^2kg^-2) M = mass causing the force m = mass the force is acting on r = distance between the centre of the two masses
only use the minus sign in front of the equation when you are interested in the direction
for example if you calculate r then the minus sign is irrelevant because you don’t need a direction
the positive direction is defined as M to m
negative is m to M
what is a field
the region in which a force operates
the stronger the gravitational field, the _______ the force on the mass
larger
relating to Newton’s law of gravitation what happens to the force if the distance between the masses doubles
inverse square law
F α 1/r^2
so if r doubles then F will be 1/4 of the original force
The gravitational force between 2 objects 10m apart is 0.291N.
What will the gravitational force between them be if they move to 25m apart?
25/10 = 2.5
inverse square law F α 1/r^2
0.291/2.5^2
= 4.7 X10^-2 N
define gravitational field strength
force acting per unit mass
what does this equation mean
g = F/m
what are the units of g
gravitational field strength = force experience by a mass in the gravitational field / mass
units of g are Nkg^-1 or ms^-2 (acceleration due to gravity)
derive the formula g = -GM / r^2
use F = - GMm / r^2 and F = mg
-GMm/r^2 = mg
cancel m
-GM/r^2 = g
sketch a graph of distance r (x) against g (y)
g is negative
rapidly increases
then decreases rapidly as distance increases
levels off
(page 66 of year 2 textbook)
(reflect in y=0 line for positive g)
2 particles of mass 0.2kg and 0.3kg are placed 0.15m apart. A third particle of mass 0.05kg is placed between them.
Calculate F acting on the third particle if it’s placed 0.05m from the 0.3kg mass.
Need to work out the two forces either side of the mass then get a resultant force.
Use equation F = GMm / r^2
F1 = 6.67X10^-11 X 0.2 X 0.05 / 0.1^2
= 6.67X10^-11 N
F2 = 6.67X10^-11 X 0.3 X 0.05 / 0.05^2
= 40.2 X10^-11 N
F = F2 - F1 = 33.5 X10^-11 N
define gravitational potential at a point
the work done in moving a unit mass from infinity to that point
what does the equation
V_g = -GM / r
mean
V_g = gravitational potential G = gravitational constant M = mass of the object causing the gravitational field r = distance from the centre of the object
At an infinite distance from the mass, the gravitational potential will be _____.
zero
What equation could you use to determine the amount of energy needed to move an object against gravity
∆W = m∆V_g
because V_g has the units of Jkg^-1 and mass is kg so when multiplying, it leaves J
what are the units of gravitational potential (V_g)
Jkg^-1
On a graph of force (y) against distance, r (x)
what does the area under the graph between 2 points equal
the work done to move the object between those two points
how are satellites kept in orbit
the gravitational ‘pull’ of the mass they’re orbitting
in our solar system the gravitational force is the centripetal force as they’re undergoing circular motion
derive an equation for orbital speed
F = GMm / r^2 and F = mv^2 / r
GMm/r^2 = mv^2/r GM/r = v^2
v^2 = GM / r
derive an equation for orbital period:
T^2 = (4π^2r^3 / GM)
v = d/t distance for a circular orbit = 2πr so v = 2πr/T T = 2πr / v
derive the equation for orbital speed to get v^2 = GM/r
sub in
T = 2πr / (√(GM/r)
square it
T^2 = (4π^2/GM)(r^3)
What are Kepler’s 3 laws of planetary motion
1) Each planet moves in an ellipse around the sun, with he sun at one focus.
2) A line joining the sun to a planet will sweep out equal areas in equal times.
3) The period of the orbit and the mean distance between the sun and the planet are related by
T^2 α r^3
A value for gravitational potential is always ___________.
negative
sketch a graph of gravitational potential (y) against radius (x)
pg 68 of year 2 textbook
what value would you get from the tangent of a gravitational potential - radius graph
what can be determined
-g
g = -∆V_g / ∆r
derive an equation for gravitational potential energy
it will be the gravitational potential multiplied by its mass
so E = mV_g = -GMm / r
what are geostationary satellites
orbit directly over the equator and are always above the same point on Earth
it travels from west to east
travels at the same angular speed as earth
has a period of 24 hours
what are the uses of geostationary satellites
communication - TV and telephone signals
monitor and track weather
If you have 2 planets with the same volume but different masses, the planet with the larger mass will have a thicker atmosphere. Why?
At any distance above its surface, the gravitational force will be greater than at the same distance above the planet with the lower mass.
So the larger mass planet can stop more atmosphere particles escaping into space, leading to a thicker atmosphere.
Two objects orbit a planet.
Object A is 4.22 X10^5 km away from the centre of the planet. Object B is 6.71 X10^5 km away from the centre of the planet.
Object A completes one orbit in 42.5 hours.
Find the orbital period of planet B, assuming both orbits are circular.
Using Kepler’s Third Law
A: 42.5^2 / (4.22X10^5)^3 = 2.403 X10^-14
that equals B: T^2 / (6.71 X10^5)^3
T^2 = 2.403X10^-14 X (6.71X10^5)^3 = 7261.194
T = 85.2 hours
= 85 hours
define escape velocity
the velocity needed so an object has just enough kinetic energy to escape a gravitational field
derive an equation for escape velocity
the kinetic energy lost = gravitational potential energy gained so
0.5mv^2 = GMm/r
cancel out m and rearrange for v
v^2 = 2GM / r
for calculating the gravitational potential energy of a satellite, why can’t you use E = mgh
because the gravitational field strength isn’t constant
