Scientific equations always need to be balanced for the most accurate possible results. Balance scientific equations with help from a research scientist and one of the world's leading experts on star formation in this free video clip.

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Scientific equations always need to be balanced for the most accurate possible results. Balance scientific equations with help from a research scientist and one of the world's leading experts on star formation in this free video clip.

Part of the Video Series: Physics & Science Lessons

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Now, there's no standard way for balancing equations. The best thing is, is you have them balance at the beginning and then, when you manipulate the equations. Whatever you do the left hand side of the equation, you have to do to the right hand side of the equation. But there is a trick to know if you're on the right track, if you've got the equation close. The force on the satellite, the gravitational force from the earth, is the gravitational constant G, capital G times the mass of the earth, divided by the distance of the satellite from the center of the earth, squared. This has units of meters per seconds squared, it's am acceleration. I want to have on the right hand side, something that gives me meters per second squared. I know that the velocity of the satellite has meters per second. So, if I square this, I now have meters squared per seconds squared. And if I then, divide by R, the distance to the center of the earth, I have now, the correct units on the right, meters per seconds squared. So, I've made sure that this equation balanced, and that's actually the two equations. Now, I can start manipulating this equation, for instance, I want to multiply both sides of the equation by R,and get rid of this R. Because I can, because, divide R into R, is one. And here, I get rid of one of these R's. So, I have GM divided by R is equal to V-squared. And this is the equation for the velocity of a satellite circling the earth. If I want to know its orbital period, I can use this same principle. The distance that a satellite goes, is the circumference of the circle of the orbit, which is two Pie times R, that's in meters. If I want to know the Period in time, I then just divide by the velocity in meters per second, and the answer is in seconds. Another example, you drop a ball from a height, H. The acceleration of gravity is again, in meters per second and the height is given in meters. So, now, if I take G and divide by H, I get something that has the units of one over seconds squared. So, so I, so the time that this ball would drop, should be something like, one over T-squared. Now, this equation is not quite right. Because what this doesn't tell me, is that there's a factor of two missing, but it almost, it almost got it right. Fourth example, let's look at the reaction of hydrogen and oxygen to produce water. Oxygen, I have two oxygen atoms here, and only one oxygen atom there. So, I have to multiply the right hand side by two, in order to get the two oxygen. Now, I have four hydrogen on the right hand side, I need four hydrogen on the left hand side. So, again, I multiply by two. And I have now balanced this chemical reaction. So, use these tricks to help you make sure that you've got the right equations.