The Gravitational Field

Transcript

0:00Introduction

Hello everyone! Okay, imagine this. We dig a tunnel straight down to the centre of the Earth, and drop a stone in. As it falls deeper, it gets closer to the centre. So gravity on it gets stronger, right? Or weaker? Keep your answer ready. Our topic: the gravitational field, how strong gravity is at every point around a body. The fields of rings, shells and spheres come up often in JEE problems. So, let's begin.

0:33The field: force per unit mass

So, what is a gravitational field? Place a small test mass, small m, at a point; gravity pulls it with a force F, okay? The field there is the force per unit mass: E equals F over m. The little arrows on top of E and F mean these are vectors, with a direction. Right? Now, the unit. Newtons over kilograms gives newton per kilogram. But some books write metre per second squared. Which one is correct? Both, yes! On screen, N kg to the power minus one is newton per kilogram; the minus one means per. And m s to the power minus two is metre per second squared. One newton is one kilogram metre per second squared, so dividing by a kilogram leaves metre per second squared. So the field is an acceleration, the small g you know. Clear? Now a point mass, capital M, at distance r from our test mass. By Newton's law, the force is capital G capital M small m, over r squared; capital G is the gravitational constant. So, what is the field there? Divide by small m, exactly! E equals capital G M over r squared. The test mass cancels; only the source mass and the distance matter. And it points towards capital M, because gravity always attracts. Got it?

1:57Ring: zero at the centre, a peak in between

Now, bodies with a size. First, a ring of mass capital M and radius capital R, okay? We take a point on its axis, the line through the centre perpendicular to the ring, at distance x from the centre. With me? The field there is capital G M x, over R squared plus x squared, in brackets, to the power three over two. Okay? Now tell me. At the very centre of the ring, what is the field? Zero, correct! Put x equals zero, and the top becomes zero. Physically, each small piece of the ring pulls one way, and the piece opposite pulls equally the other way. They cancel in pairs. Right? So, zero at the centre. And far along the axis, the bottom grows faster than the top, so the field is small again. So in between, it must have a peak. Where? At x equals R over root two: capital R over the square root of two. There, E max, with max written small below the E, equals two capital G M, over three root three R squared. Clear?

2:59Thin spherical shell

Next, a thin spherical shell: a hollow ball, its mass capital M spread on a surface of radius capital R, and small r is the distance from its centre. Imagine you float inside, off the centre. Which way does the shell pull you? Nowhere at all! Inside, where r is less than capital R, the field is exactly zero; that sign means less than. Not just at the centre, but everywhere inside. Students often forget this. Okay? And outside, where r is greater than capital R, E equals capital G M over r squared: the same as a point mass at the centre. Got it?

3:37Uniform solid sphere

Now the most important one: a uniform solid sphere, like a planet. Look at the graph. Along the bottom, r, the distance from the centre; going up, small g, or the field E, okay? The dashed line is at r equals capital R, the surface: inside on its left, outside on its right. With me? Inside, E equals capital G M r over R cubed. Capital R is fixed, so E is proportional to r; this curly sign means proportional to. That is the straight line rising from zero, marked small g proportional to r. Okay? Outside, E equals capital G M over r squared, just like a point mass. That is the falling curve, marked small g proportional to one over r squared. Right? So, where is the field the greatest? At the surface, exactly! The peak is at r equals capital R, and its height is marked g nought on the vertical axis; the small zero below the g is a subscript. Put r equals capital R in either formula: both give capital G M over R squared. Clear? Now, our tunnel. Treat the Earth as a uniform solid sphere. Deeper means closer, so stronger? No, no, careful! Inside, the field is proportional to r, so it gets weaker as the stone goes deeper, and it is zero at the centre. Closer means stronger only outside. Got it?

5:05Uniform rod, on its axis

One last body: a uniform rod, mass capital M, length capital L, and a point on its axis, along the rod's own line, at distance d from the near end. Okay? Now, a quick trap. Can we treat the rod as a point mass at distance d, and write capital G M over d squared? No, careful! Its far parts are farther away, and pull less. The correct field is capital G M, over d times, d plus capital L, in brackets. And d is measured from the near end, not from the middle. Right?

5:38Example 1: field of a shell at R/2 and at 2R

Example one, from your course. Take a uniform shell of mass capital M and radius capital R, okay? Find the field at R over two, and at two R, from its centre. Pause the video, and try it first. Done? Okay, let's check. R over two is less than capital R, so this point is inside the shell, and inside a shell the field is zero. No calculation at all! Right? Next, two R: outside, so the shell acts like a point mass at its centre. E equals capital G M over two R, in brackets, squared, which is capital G M over four R squared. Careful: square the two as well, so four, not two. Got it?

6:23Example 2: solid sphere, two points with 1/4 surface field

Example two, a little harder. At what two distances from the centre of a uniform solid sphere is the field one quarter of its surface value? Use the graph. Pause the video, and try it. Alright. Outside, the field goes as one over r squared. So one quarter of the field means double the distance: r equals two R, done, right? No, no, wait! The question says two distances. On the graph, the field reaches one quarter of the peak twice: once on the rising line, once on the falling curve. First, the surface value: E s, with a small s below for surface, equals capital G M over R squared. Okay? Inside, the ratio E over E s: capital G M r over R cubed, divided by capital G M over R squared, is simply r over capital R, nothing more. Set it equal to one quarter; the double arrow means so: r equals R over four. In our tunnel, that point is three quarters of the way down! Clear? Outside, the ratio is capital R squared over r squared, equal to one quarter, so r equals two R, as we guessed. Two answers, one on each side of the peak: R over four inside, and two R outside. See how the graph saved us?

7:44Quick revision: field of point mass, ring, shell, sphere

So, quick revision. The field is force per unit mass, and for a point mass it is capital G M over r squared. Ring: zero at the centre, greatest at R over root two. Shell: zero inside. Solid sphere: proportional to r inside, greatest at the surface. Now go and try the practice problems on this page. You can do them. All the best!