Escape Velocity

Transcript

0:00Introduction

Hello everyone! Okay, try this in your head. Throw a ball straight up. It rises, stops, and comes back down. Throw it harder, and it goes higher, but it still comes back, right? So, is there a speed so large that it never comes back at all? Yes, there is, and today we find it. And here is the real puzzle: does a huge rocket need a bigger speed than a small pebble? Most students say yes. By the end of this video, you will know exactly why. Escape velocity comes up often in JEE gravitation. So, let's begin.

0:39Escape velocity: just enough energy to reach infinity

So, what does never coming back really mean? It means the body goes all the way out to infinity. We take the Earth as a spherically symmetric body, of mass capital M and radius capital R, and we ignore air for now, okay? Now the energies. At the surface, a body of mass small m has potential energy capital U equals minus capital G M m over R; capital G is the gravitational constant. And at infinity, as we saw earlier in this chapter, capital U is zero. Right? So, how much kinetic energy must we give it at the surface, just enough to reach infinity, and not a bit more? Capital G M m over R, yes, exactly! Just enough means it reaches infinity with zero speed, so the total energy is zero at both ends. So, half m v e squared equals capital G M m over R; v e, with a small e written below, is the escape velocity. With me? Cancel small m, and v e equals the square root of two capital G M over R; the tick-shaped sign is the square root. And capital G M equals small g R squared, from the last chapter, so this is also root of two g R, the form you will use most. Clear? Now the Earth's numbers: small g is nine point eight metres per second squared, and R is six point four times ten to the power six metres. Two times nine point eight times six point four times ten to the six, then the root. The wavy equals sign means approximately: about eleven point two kilometres per second; the s to the minus one just means per second. Remember this number. Got it?

2:17Same for a pebble and a rocket

Now, back to our puzzle. Think about the formula we just found. Does a huge rocket need a bigger escape velocity than a small pebble? No! Small m cancelled, remember? So the escape velocity is the same for the pebble and the rocket. Puzzle solved! The rocket needs far more energy, of course, but not more speed. Okay? And the direction? Straight up or at an angle, it does not matter. Energy has no direction; only the speed enters the kinetic energy. One condition, though: we are ignoring air. Right? Now one more useful form. Write the mass as density times volume: four thirds pi R cubed, times rho; rho, the Greek letter that looks like a p, is the density. Put it in, and v e squared becomes eight thirds pi capital G rho R squared; the double arrow means implies. Clear? So, v e is proportional to R times root rho; the little fish-shaped sign means proportional to. This form is perfect when a question compares planets of the same density. With me?

3:22Slower than escape velocity: how high does it go?

Now, what if we throw it straight up, but slower than v e? The less-than sign says v is less than v e. It rises to some height h, stops there, and falls back. Okay? At the top, all its kinetic energy has gone into lifting it. So per kilogram, half v squared equals g h over, one plus h over R; we found this lifting work earlier in this chapter. Multiply the top and the bottom by R, and it is g R times h over, R plus h. Right? Now rearrange for h, using v e squared equals two g R, and you get this. h equals R, over, v e over v, the whole squared, minus one. One line, no need to redo the energy every time. Got it? Now let's test it. What if v is exactly equal to v e? What does this formula give?

4:14Faster than escape velocity: speed left at infinity

Infinity! Yes, exactly! The bottom becomes one minus one, which is zero, so h shoots up to infinity; the arrow means tends to, and the sideways eight is the infinity sign. The body just escapes, so the formula agrees with our definition. Clear? Now go faster than v e; that is the greater-than sign. Half v e squared per kilogram is spent in escaping. Whatever is left stays as kinetic energy far away: half v infinity squared, with the infinity sign written below the v. Okay? So the speed far away is just v minus v e, right? No, no, wait! That is a common trap. Energies subtract, not speeds. Cancel the halves, and v infinity equals the root of, v squared minus v e squared. Got it?

5:04Example 1: same density, twice the radius

Now, the first example. Here is a planet with the same density as the Earth, and twice its radius. Find its escape velocity. Pause the video, and try it first. Done? Okay, let's check. Bigger radius, so the surface is farther from the centre. R sits in the bottom of the formula, so the escape velocity goes down, right? Oh no, careful! The mass does not stay the same. Use the density form: v e is proportional to R root rho. Rho is the same and R doubles, so v e doubles. Okay? Let's double-check with the mass. Mass goes as R cubed, so twice the radius means eight times the mass. Eight capital M over two R is four times capital M over R, and the root of four is two. Same answer. Right? So, two times eleven point two: twenty two point four kilometres per second. The mass grows with the radius; never forget that. That's it!

6:03Example 2: height reached at half the escape velocity

Now a harder one, JEE style. Here, a body is projected straight up from the Earth at half the escape velocity. How high does it rise? Pause and try it. Hint: does small g stay the same all the way up? Done? Okay, let's check. Quick way: h equals v squared over two g, the formula from projectile motion. Easy, right? No, no, wait! That is the trap. Here v squared is a quarter of v e squared, so two g R over four; divide by two g, and you get R over four. But that formula assumes g stays constant, and it does not: high up, g is weaker, so the body climbs higher than that. Okay? So, use our formula. v e over v is two, and two squared is four. h equals R over, four minus one: R over three. Clear? With R as six thousand four hundred kilometres, h is about two thousand one hundred and thirty three kilometres. See? One third of R, not a quarter. That's it!

7:06Quick revision: escape speed, height reached, speed left

So, quick revision. Escape velocity is root of two capital G M over R, which is root of two g R, about eleven point two kilometres per second for the Earth, and it goes as R root rho. It is the same for any mass and any direction, ignoring air. Slower than v e, straight up: h equals R over, v e over v, the whole squared, minus one. Faster: v infinity is root of, v squared minus v e squared. Now go and try the practice problems on this page. All the best!