Newton's Law of Gravitation

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Newton said that every particle in the universe attracts every other particle. So this pen attracts this book, and you attract the friend sitting next to you! Then why don't you feel your friend pulling you? Hold on to your answer. Today we learn Newton's law of gravitation, and we will solve this puzzle too. Gravitation questions come up often in JEE, and they all build on this law. So, let's begin.

0:32Newton's law of gravitation

So, here is the law. Every particle attracts every other particle: every pair, anywhere. Okay? Now, how strong is the pull? Capital F, the force between two particles, equals capital G times m one times m two, over r squared. The small subscripts one and two label the two masses, and r is the distance between them. Capital G is the gravitational constant. Clear? So the force is proportional to the product of the masses; the open loop sign means proportional to. And it is inversely proportional to the square of the distance, so F is proportional to one over r squared. An inverse square law. Right? Now tell me. If I double the distance between the two masses, the force becomes half, right? No, no, careful! The distance is squared. Two squared is four, so the force becomes F over four, not F over two. Don't fall for this inverse square trap. Got it?

1:32The gravitational constant G

Now, capital G itself. Rearrange the law: G equals F times r squared, over m one times m two. So, what is the unit of G? Yes, exactly! Newton times metre squared, over kilogram squared. So G equals six point six seven times ten to the minus eleven newton metre squared per kilogram squared; the small minus eleven up top is the power of ten, and kilogram to the minus two means per kilogram squared. Really tiny! Okay? And its dimensions. The square brackets mean dimensions: M for mass, L for length, and T for time. One newton is MLT to the minus two; multiply by L squared, and divide by M squared. That gives M to the minus one, L cubed, T to the minus two. With me?

2:21The law as a vector

Force has a direction, so let's write the law as a vector. The little arrow on top of a letter means a vector, and the small one two written below is a subscript. Vector F one two equals minus G m one m two over r cubed, times vector r one two. Okay? Vector r one two points from m one to m two. And vector F one two is the force on m two. Clear? Now, why the minus sign? Vector r one two points away from m one, towards m two. So which way does the force on m two point? Back towards m one, yes! The minus sign reverses the direction of r one two. That is attraction, written in vector language. Right? But wait. At the bottom it says r cubed, not r squared. So has the law become an inverse cube law? No, no, careful! The two straight bars around a vector mean its magnitude, its length. Vector r one two has length r, so one r on the bottom cancels with it. The size of the force is still G m one m two over r squared. Same law, now with a direction. Got it?

3:32Features of the gravitational force

Now, the features of this force. First, it is always attractive. Gravity never pushes; it only pulls. And it acts along the line of centres. Okay? Now, the Earth pulls an apple. Does the apple also pull the Earth? With how much force? Yes, exactly! With the same force! The two forces form an action reaction pair: equal in size, opposite in direction. The Earth hardly moves only because its mass is so huge. Clear? Third, the medium does not matter. Air, water, or anything else between the bodies: the force stays the same. With me? Fourth, superposition. If many masses pull on one particle, find the force due to each mass separately. Then add them as vectors, not as plain numbers. Right? And last, a very useful one. Take a uniform sphere, or a spherical shell, and a particle outside it. The sphere attracts the particle as if all its mass were at its centre. So, for that outside particle, measure r from the centre of the sphere. But remember the conditions: uniform, and the particle outside.

4:45Example 1: force between two 1 kg masses 1 m apart

Okay? Example one, an easy one. Find the gravitational force between two one kilogram masses, one metre apart. Then compare it with the weight of one kilogram. Pause the video, and try it. Done? Okay, let's check. Put m one and m two equal to one kilogram, and r equal to one metre. Then F is G times one times one, over one squared: just G itself! Six point six seven times ten to the minus eleven newton. Clear? Now the weight of one kilogram. Weight is m times small g, the acceleration due to gravity near the Earth's surface. One times nine point eight: nine point eight newton. Right? So, how many times larger is the weight? Divide nine point eight by six point six seven times ten to the minus eleven. The wavy equals sign means approximately: about one point five times ten to the eleven. Okay? So the weight is about one hundred and fifty billion times larger! Now, remember our puzzle? You do attract your friend, but between everyday bodies this pull is incredibly tiny. We feel gravity strongly only when one of the bodies is huge, like the Earth. Perfect!

6:09Example 2: three equal masses on an equilateral triangle

Now example two, a little harder. Three particles, each of mass m, sit at the corners of an equilateral triangle of side a. Find the net force on one of them. Pause the video, and try it. Hint: superposition. Done? Okay, let's check. Pick any one particle. The other two each pull it along a side, towards themselves. Each pull is F nought, F with a small zero: G m squared over a squared, since both masses are m and the distance is a. Clear? So, two pulls of F nought each. The net force is two F nought, right? Oh no, careful! Forces are vectors. In an equilateral triangle every angle is sixty degrees; the small circle up top means degrees. So the two pulls are sixty degrees apart, and you cannot just add their sizes. With me? Draw the line that cuts that sixty degree angle in half: the bisector. Each pull makes thirty degrees with it. The sideways parts, across the bisector, are equal and opposite, so they cancel. Okay? The parts along the bisector add up. Each is F nought times cos thirty degrees, so F equals two F nought cos thirty degrees. Cos thirty is root three over two; the tick shaped sign is the square root, and the dots mean times. So F equals two, times G m squared over a squared, times root three over two. Right? The twos cancel. F equals root three times G m squared, over a squared. And the direction? Along the bisector, which passes through the centroid, the centre of the triangle. So the net force points towards the centroid. Not two F nought, but root three F nought. Perfect!

7:55Quick revision: F = Gm1m2/r^2, value of G, superposition

So, quick revision. Capital F equals G m one m two over r squared: always attractive, and inverse square. Capital G is six point six seven times ten to the minus eleven newton metre squared per kilogram squared, so gravity between small bodies is tiny. With many masses, add the forces as vectors; and outside a uniform sphere, measure from its centre. Now go and try the practice problems on this page. All the best!