Gravitational Fields IB Physics Power Points Topic 6 SL Fields and Forces

Defining Mass Inertial mass A measure of an object’s inertia (tendency to resist a change in motion) The property of matter that gives rise to the gravitational attraction between objects. The gravitational force is proportional to the product of the two object’s masses. Gravitational mass Expressed as a ratio of net force and acceleration – Newton’s 2 nd Law

Newton’s Universal Law of Gravitation Consider – two masses A and B This gravitation attraction can be calculated by There is a gravitational attraction between A and B. F B on A = - F A on B (think Newton’s 3 rd )

The gravitational constant G = 6.67× N m 2 kg -2 The gravitational force is: Proportional to the product of the two masses. Inversely proportional to the square of the distance between the center of mass of each object. NULOG  “an inverse square law”

How could we determine G experimentally?

Practice: Find the gravitational force of attraction between a 75 kg physics student and her kg car when their centers are 10.0 m apart. A.7.5 x10 -8 N B.7.5 x10 -7 N C.740 N D.1.5 x10 3 N

Falling objects and the force of gravity

Free-body diagrams and elevators – what do scales measure?  Draw free body diagrams (including annotations and calculated values) representing the following situations (consider g = 10 ms -2 ) i.A 50 kg student in an elevator at rest ii.A 50 kg student in an elevator moving downwards at 2 ms -1. iii.A 50 kg student in an elevator moving upwards at 5 ms -1. iv.A 50 kg student in an elevator accelerating upwards at 1 ms -2. v.A 50 kg student in an elevator accelerating downwards at 3 ms -2 vi.A 50 kg student in an elevator falling freely

Consider an object (mass m) at some point P, in the Earth’s gravitational field. The object will feel a force (F e ) that can be calculated using:

Gravitational Field Strength Defined – the force per unit mass acting on an object placed in a gravitational field. In the example, the force on the object at point P in the Earth’s gravitational field can be calculated by:

So by substitution The gravitational field strength (FORCE PER UNIT MASS) at distance r from the center of mass M (the object creating the gravitational field)

At the surface of a planet (like Earth), the gravitational field strength can be determined by: Where M is the mass of the planet. For Earth:

Given Earth’s mass: 5.97 x kg Earth radius: ~ 6.38 x 10 3 km Determine a value for the gravitational field strength at the Earth’s surface.

Gravitational Fields Gravity is a field force. It exists between objects without contact and cannot be shielded or blocked. Gravitational fields are depicted by drawing field diagrams. Field lines show the direction of the force acting on a particle placed in the field (toward the center of mass) The density of field lines indicates the strength of the field.

Gravitational Field Strength – More than one point mass Consider the following diagram and determine the gravitational field strength at point P, located 3.45 x 10 8 m from the centre of the Earth. Moon mass: 7.35 x kg Earth-Moon distance: 384 x 10 3 km

The gravitational fields of the earth and moon superpose. Note how the fields cancel at one point, and how there is no boundary between the interpenetrating fields surrounding the two bodies.