# 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) |

r^{2} |

**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 / r

^{2}Combining, magnitude of g = GM / r

^{2}Therefore

g = GM / r, M = Mass of object “creating” the field^{2}

##### Example 1:

Assuming that the Earth is a uniform sphere of radius 6.4 x 10^{6} m and mass 6.0 x 10^{24} kg, find the gravitational field strength g at a point:

(a) on the surface,

g = GM / r

^{2}= (6.67 × 10^{-11})(6.0 x 10^{24}) / (6.4 x 10^{6})^{2}= 9.77ms^{-2}(b) at height 0.50 times the radius of above the Earth's surface.

g = GM / r

^{2}= (6.67 × 10^{-11})(6.0 x 10^{24}) / ( (1.5 × 6.4 x 10^{6})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 / r

^{2}

gP / gE = M_{P}r_{E}^{2}/ M_{E}r_{P}^{2}= (4/3) π r_{P}^{3}r_{E}^{2}ρ_{P}/ (4/3) π r_{E}^{3}r_{P}^{2}ρ_{E}= r_{P}/ r_{E}= 2Hence g

_{P}= 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 / r^{2}. When a particle is released, it will fall towards the centre of the Earth, as a result of the gravitational force with an acceleration a_{g}.

F_{G} = ma_{g}

a_{g} = GM / r^{2}

Hence a_{g} = 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 g_{o}, express its apparent weight, N, of a body of mass m in terms of m, g_{o}, 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 = GM

_{E}m / R^{2}= mg^{o}At the equator,

Normal Reaction Force on ship by Earth = Gravitational attraction - centripetal force

N = mg_{o}– mRω^{2}= mg_{o}– 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 = mφ _{f}-mφ_{i}**

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: F _{e} = Q_{1}Q_{2} / 4πε_{o}r^{2} |
Newton's Law of Gravitation: F _{g} = G (GMm / r^{2}) |

4. | Field Strength of isolated Point Charge or Mass | E = Q / 4πε_{o}r^{2} |
g = G (GM / r^{2}) |

5. | Definition of Potential | Work done in bringing a unit positive charge from infinity to the point; V = W /Q | Work done in bringing a unit mass from infinity to the point; φ = W / M |

6. | Potential of isolated Point Charge or Mass | V = Q / 4πε_{o}r |
φ = -G (M / r) |

7. | Change in Potential Energy | ΔU = q ΔV | ΔU = m Δφ |

Total Energy of a Satellite = GPE + KE = (-GMm / r) + ½(GMm / r)

##### Escape Speed of a Satellite

By Conservation of Energy,

Initial KE | + | Initial GPE | = | Final KE | + | Final GPE |

(½mv_{E}^{2}) |
+ | (-GMm / r) | = | (0) | + | (0) |

Thus escape speed, v_{E} = √(2GM / R)

Note : Escape speed of an object is independent of its mass

For a satellite in circular orbit, "*the centripetal force is provided by the gravitational force*"

**{Must always state what force is providing the centripetal force before following eqn is used!}**

###### Hence GMm / r^{2} = mv^{2} / r = mrω^{2} = mr (2π / T)^{2}

A satellite does not move in the direction of the gravitational force {ie it stays in its circular orbit} because: the gravitational force exerted by the Earth on the satellite is __ just sufficient to cause the centripetal acceleration but not enough to also pull it down towards the Earth__.

{This explains also why the Moon does not fall towards the Earth}

Geostationary satellite is one which is __always above a certain point on the Earth__ (as the Earth rotates about its axis.)

For a **geostationary** orbit: T = 24 hrs, orbital radius (& height) are fixed values from the centre of the Earth, ang velocity w is also a fixed value; rotates fr west to east. However, the __mass__ of the satellite is __NOT a particular value__ & hence the ke, gpe, & the centripetal force are also not fixed values {ie their values depend on the mass of the geostationary satellite.}

A geostationary orbit must lie in the ** equatorial plane** of the earth because it

__must__accelerate in a plane where the

*centre*of Earth lies since the

__exerted on the satellite is the__

*net*force__Earth's gravitational force__, which is

__directed towards the__.

*centre*of Earth{Alternatively, may explain by showing why it's impossible for a satellite in a non-equatorial plane to be geostationary.}