Kepler’s 3rd Law ( T 2 ∝ r 3 ) Gravitational force provides centripetal force

mrω 2 =

GMm

r 2⇔

2π

T

⎛⎝⎜

⎞⎠⎟

2

=GM

r 3⇔ T 2 =

4π 2

GMr 3

Law of Gravitation The force of attraction between any two point masses is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

Fgrav and field strength g

Gravitation field strength g Gravitational field strength at a point is defined as the gravitational force per unit mass acting on a small test mass placed at that point.

g = F

m

GRAVITATIONAL FORCE AND FIELD STRENGTH

Gravitational field near surface of planet

• Earth’s curvature negligible at this scale • Approx. uniform g-field

o field lines near the surface ⊥ to surface o equal separation lines ⇒ constant g

Questions you should know how to solve 1. Use law of gravitation and field strength formula for up to a few bodies (include

the use of vector addition). 2. Determine point with zero field strength between 2 masses. 3. Compare field strengths given ratio of radius/density/mass. 4. Circular motion with gravitational force provides the centripetal force. 5. Use Kepler’s 3rd law to compare periods of orbits. 6. Recognise that the balance reading is dependent on the position of

measurement due to earth’s self-rotation.

Balance readings at different positions by considering earth’s self rotation

!

Satellites • Use circular motion concepts (see LHS) • Geostationary – T=24h, above equator at

fixed distance from earth, moves in same direction as earth’s rotation

                                                       

             

GRAVITATIONAL POTENTIAL AND POTENTIAL ENERGY

M (fixed)

m

r

FG

∞

Fext

final position

initial position

Gravitational potential energy …is the work done by an external agent in bringing the mass from infinity to its present location (without any change in kinetic energy). For 2 point masses:

Ugrav = −GMm

r unit: J( )

Expression is negative: 1. Grav. PE is zero at infinity. 2. External force is opposite to the

displacement of the mass m. 3. W.D by external agent is -ve.

Gravitational potential Gravitational potential at a point is defined as the work done in bringing unit mass from infinity to the point.

φ =Ugrav / m Note: the mass m refers to the mass that is being moved For a point mass

φ = −GM

r unit: J kg−1( )

Solving problems using Ugrav and Φ

(Step 1) Δ φ between two points

Δφ = φfinal − φinitial = φB − φA (Step 2) change in gravitational potential energy of mass m

ΔUgrav = mΔφ  Caution! Negative sign of φ must be substituted Caution! When calculating the change in potential energy of the mass, the positive or negative sign of the potential difference must be substituted ΔUgrav Interpretations Positive • Potential energy of mass increases

o External work done/energy transfer without change in KE, or,

o Loss in KE

Negative • Potential energy of mass decreases o Energy transfer out without change in

KE, or, o Gain in KE

Field strength and potential Gravitational field strength at a point is numerically equal to the potential gradient at the point.

g = −

dφdr

or graphically, it is the gradient ΔφΔr

⎛⎝⎜

⎞⎠⎟

Note: -ve indicates g is in direction of decreasing φ

!

Questions you should know how to solve 1. Compute potential (or GPE) at a point (or mass) due to a few bodies (using scalar addition). 2. Compute change in potential (or GPE) between final and initial position (of a mass). 3. Obtain the field strength at a point from the graph of potential with position for simple situations.

4. Understand total energy ET = EK + Ugrav and its implication and for object/satellite in orbit ( E

K=

GMm

2r U

grav= −

GMm

2r E

T= −

GMm

2r)

5. Use conservation of energy (ET,initial = ET, final or ΔUgrav = ΔEK) for different situations a) motion between two points b) escape velociy