… Gravitational fields are always attractive.

… Newton’s law of universal gravitational: the attractive force between two objects separated by a distance r is given by:

F = GMm/r²

G is the Universal Gravitational Constant = 6.67×10^-11 Nm²/kg²

… Gravitational field strength, g, is the force experienced per unit mass (N/kg)

g = F/m , units N/kg

For radial fields:

g = GM/r²

… Gravitational field lines around a body of mass M show the direction of the gravitational force that acts on other masses, around it. These lines always point inwards towards the centre of mass of the body.

The more densely spaced the lines are, the stronger the gravitational field is at that point.

In an approximately uniform field (for example, close to the earth’s surface), gravitational field lines are parallel.

Gravitational field lines can be ‘warped’ by other masses that are nearby. For example, here is a diagram of the gravitational field lines between the Earth and the Moon: https://goo.gl/images/ZzRgM7

… Gravitational potential energy, Ep, is measured from a reference point of r = ∞ where Ep = 0J. This means that as a mass falls towards the Earth, its Ep gets more and more negative!

Ep = -GMm/r

… The change in gravitational potential energy by lifting a mass from a distance r to R:

ΔEp = GMm(1/R – 1/r)

… Gravitational POTENTIAL, Vg, is similar to gravitational potential energy, but is on a ‘per kg’ basis. So when you know the potential at a point, you can work out the potential energy of any object at that point:

Ep = Vg × m

At r=∞, Vg = 0J/kg

Vg = -GM/r

… To work out the escape velocity of a mass:

find the gravitational potential energy of the mass at its present position (a negative quantity)

Kinetic energy can be given to the mass to move it infinitely far away (where the gravitational potential energy will be 0J).

Use ½mv² = Ep to work out v, the escape velocity.

… The gradient of a graph of Ep vs r equals the force acting on the mass being considered:

d(Ep)/dr = -F

… The gradient of a graph of Vg vs r equals the field strength at that point:

d(Vg)/dr = -g

… Lines of ‘equipotential’ show the gravitational potential energy of a unit 1kg mass at that point. These lines are circular around the Earth, but get further apart as r increases.

In an approximately uniform field (e.g. close to the Earth’s surface), these lines of equipotential are evenly spaced.

… Note that lines of equipotential are always at right angles to the field lines.

… To find the velocity (speed) of a body which is orbiting a planet in an approximately circular orbit:

centripetal force = gravitational force

mv²/r = GMm/r²

so

v = √(GM/r) … Note that orbit speed does NOT depend on the mass of the object!

… To find the angular velocity, ω, of a body orbiting a planet:

centripetal force = gravitational force

mrω² = GMm/r²

ω = √(GM/r³)

… By using distance = speed x time, we can find the time period of orbit:

circumference of orbit = orbit speed x Period

2πr = √(GM/r) × T

so

T² = 4π²r³/GM … which is known as Keplar’s Third Law (T²/r³ = constant)

Geostationary (geosynchronous) orbit

… Is only possible at a particular radius above the equator. So this means there is only one ‘ring’ in space around a planet which is available to place a geostationary satellite.

… For geostationary orbit, a satellite must remain above a fixed point on the Earth’s surface. So the angular velocity of the the satellite must equal the angular velocity of the earth’s rotation,

ω = 2π/T = 2π/(24×3600)

The gravitational force towards the planet must be exactly the right magnitude to supply the centripetal force to keep the object in circular motion:

Centripetal force required for circular motion = gravitational force available

mrω² = GMm/r²

r³ = GM/ω²

… A ratio approach can be useful when you are given a comparison between two quantities. For example:

“The Earth is 81 times more massive than the Moon. The gravitational field strength at the surface of the Earth is only 6 times greater than that at the surface of the Moon.

Use the above data to compare the radius of the Earth with that of the Moon.”

Here, start by writing down a relevant equation which in this case is g = GM/r² for both the Earth and the Moon. (So ‘Me’ would be the mass of the Earth etc).

Then divide one equation by the other – you’ll find that any constants will cancel out.

Finally rearrange to make the subject the ratio you are interested in.

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Earth-Moon Zero point

… What is the gravitational potential at the ‘zero point’ between the Earth and Moon where the gravitational field = 0?

The curve of potential vs distance does NOT touch the axis (so Vg ≠ 0J/kg)…

The Vg at a point measures the amount of energy required to bring an object of mass 1kg from infinity, where the value is zero, to that point.

As gravity is an attractive force, you have have to do work to move away from another object. This means that you have to give it energy to get its value to zero at infinity. Therefore the values near the Earth or Moon must be negative – you have to add energy to get the value to zero.

Imagine moving the 1kg away from the ‘zero point’ to a few million km away… So you could look back and see the Moon and the Earth as two little balls quite ‘close’ to each other.

As you move further away, you would still need to do work against the attractive forces of the moon and the earth pulling you towards them. There is no path of zero force that leads from the ‘zero point’ between the Earth and Moon to infinity… And so the potential at the zero point must still be negative (although a local minimum point on the curve).

Here is more about that specific question:

http://www.cyberphysics.co.uk/Q&A/KS5/gravitation/gravitationA5.html