# Physics Notes

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• Published : April 20, 2013

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Gravitation
Gravitational field strength at a point is defined as the gravitational force per unit mass at that point. Newton's law of gravitation:
The (mutual) gravitational force F between two point masses M and m separated by a distance r is given by F =| GMm| (where G: Universal gravitational constant)|
| r2| |
or, the gravitational force of between two point masses is proportional to the product of their masses & inversely proportional to the square of their separation. Gravitational field strength at a point is the gravitational force per unit mass at that point. It is a vector and its S.I. unit is N kg-1. By definition, g = F / m

By Newton Law of Gravitation, F = GMm / r2
Combining, magnitude of g = GM / r2
Therefore g = GM / r2, M = Mass of object “creating” the field Example 1:
Assuming that the Earth is a uniform sphere of radius 6.4 x 106 m and mass 6.0 x 1024 kg, find the gravitational field strength g at a point: (a) on the surface,
g = GM / r2 = (6.67 × 10-11)(6.0 x 1024) / (6.4 x 106)2 = 9.77ms-2 (b) at height 0.50 times the radius of above the Earth's surface. g = GM / r2 = (6.67 × 10-11)(6.0 x 1024) / ( (1.5 × 6.4 x 106)2 = 4.34ms-2 Example 2:

The acceleration due to gravity at the Earth's surface is 9.80ms-2. Calculate the acceleration due to gravity on a planet which has the same density but twice the radius of Earth. g = GM / r2
gP / gE = MPrE2 / MErP2 = (4/3) π rP3rE2ρP / (4/3) π rE3rP2ρE = rP / rE = 2 Hence gP = 2 x 9.81 = 19.6ms-2
Assuming that Earth is a uniform sphere of mass M. The magnitude of the gravitational force from Earth on a particle of mass m, located outside Earth a distance r from the centre of the Earth is F = GMm / r2. When a particle is released, it will fall towards the centre of the Earth, as a result of the gravitational force with an acceleration ag. FG = mag

ag = GM / r2
Hence ag = g
Thus gravitational field strength g is also numerically equal to the acceleration of free fall. Example 1:
A ship is at rest on the Earth's equator. Assuming the earth to be a perfect sphere of radius R and the acceleration due to gravity at the poles is go, express its apparent weight, N, of a body of mass m in terms of m, go, R and T (the period of the earth's rotation about its axis, which is one day). At the North Pole, the gravitational attraction is F = GMEm / R2 = mgo At the equator,

Normal Reaction Force on ship by Earth = Gravitational attraction - centripetal force N = mgo – mRω2= mgo – mR (2π / T)2
Gravitational potential at a point is defined as the work done (by an external agent) in bringing a unit mass from infinity to that point (without changing its kinetic energy). φ = W / m = -GM / r

Why gravitational potential values are always negative?
As the gravitational force on the mass is attractive, the work done by an ext agent in bringing unit mass from infinity to any point in the field will be negative work {as the force exerted by the ext agent is opposite in direction to the displacement to ensure that ΔKE = 0} Hence by the definition of negative work, all values of φ are negative. g = -| dφ| = - gradient of φ-r graph {Analogy: E = -dV/dx}| | dr| |

Gravitational potential energy U of a mass m at a point in the gravitational field of another mass M, is the work done in bringing that mass m {NOT: unit mass, or a mass} from infinity to that point. → U = m φ = -GMm / r

Change in GPE, ΔU = mgh only if g is constant over the distance h; {→ h<< radius of planet} otherwise, must use: ΔU = mφf-mφi
| Aspects| Electric Field| Gravitational Field|
1.| Quantity interacting with or producing the field| Charge Q| Mass M| 2.| Definition of Field Strength| Force per unit positive charge E = F / q| Force per unit mass
g = F / M|
3.| Force between two Point Charges or Masses| Coulomb's Law: Fe = Q1Q2 / 4πεor2| Newton's Law of Gravitation:
Fg = G (GMm / r2)|
4.| Field Strength of isolated Point Charge or Mass| E = Q / 4πεor2...