Below is the written transcript of my YouTube tutorial video MCAT Physics Forces – Gravitational Forces in MCAT Physics

Click the image to watch the video or catch the entire MCAT Physics Forces series HERE

(click here to see the video on YouTube)

**[START TRANSCRIPT]**

Leah here from Leah4sci.com/MCAT and in this video we'll look at the Force of gravity as it shows up on your MCAT with you at the surface of the earth and for objects on the outer space for example a satellite that is orbiting the earth.

You can find my entire video series on Forces alongwith my practice quiz and cheatsheet by visiting my website leah4sci.com/Forces.

When you think of gravitational forces, think of it as the force of gravity that pulls you to the center of the earth. This is something that you can visualize because if you drop something it tends pull down, if you jump up, once you reach the top you will turn around and fall back down. If force is equal to mass times acceleration, the force of gravity has to fit this format and so we get that the force of gravity is equal to mass times the acceleration of gravity or simply m times g. The force of gravity changes depending on how close or far you are from the surface of the earth. But if you think of the regular wall that we live in where you walk, ride a bike, that is considered the general surface that has a gravitational acceleration that is approximately 9.81 meters per seconds squared. But you can't do calculations for this number on the MCAT without a calculator so you wanna estimate the acceleration of gravity, it'll be approximately ten meters per second squared making it very easy to apply the calculations.

And notice that the units are meters per second squared just like acceleration. But don't just memorize that F equals mg, I wanna make sure you understand how to apply this no matter which aspect of the equation you're given. So with F is equal to mg and you're asked to solve for mass, simply divide both sides by g and mass is equal to Force over Gravity. Ones again, if F is mass times gravity, and this time you're asked to solve for gravity, and let's say you're given a force and you're told that gravity differs from the 10 meters per second squared, just divide both sides by m and your new gravitational constant is equal to F over M. Since gravitational force holds an object towards the center of the earth and that's calculated as mass times gravity we can also think of that as weight. The weight is the pull that you feel towards earth and that's why you have a certain weight on earth but if you go to one of the planet, you'll have a different weight because there's different gravitational constant pulling you to that planet.

The weight of an object is simply equal to mass times the acceleration of the gravity which as we already know is the force of gravity. But gravity isn't limited to just an object being pulled to the center of the earth. Any two objects can have a gravitational pull on each other. The reason we typically don't calculate that is the force of gravity towards earth is much stronger than the gravity of two objects attracting each other. Now what happens if you're far away from the center of the earth such as in an airplane flying really high or even in a spaceship flying in an outer space. We can no longer use g is equal to 9.81 so we have to use a different equation and the equation we use is the force of gravity is equal to g, big M little m over r squared where g is the gravitational constant of 6.7 times ten to the minus eleven newton meter square per kilograms square. Big M is the mass of the planet in question, little m is the object on that planet and r squared is the square of the distance between them from the center of the object to the center of the planet. So if we compare this equation to the simple fg is equal to mass times gravity, the mass is still there in the equation and that means everything that's left over is estimated to equal ten meters per second square at the surface of the earth. In other words, g is equal to big G, big M over r squared.

For example, if I look at the circle to represent earth and then I have a small object orbiting earth, we have to count the radius as the distance from the center of earth center of the object and that right there will be the r value. In addition to memorizing this equation, you also want to understand what we're looking at because you may find yourself with the question, that doesn't ask for calculation but only asks for a proportion base on logical understanding. For example, if you see a question that says;

**Find the gravitational force of a person of a person standing on the surface of a planet that has half the mass and twice the radius of earth.**

And you look at choices but instead of numbers, you're given some value and then the letter f. What this tells you is we're looking at how the force of gravity will change when the person is standing on a planet rather than the planet earth. And the way we set this up is simply by writing out the equation for the force of gravity on earth and comparing it to the new force of gravity. So the force of gravity on earth is equal to G big M little m over r squared and everyt unit here is one. So we have one times the mass of earth, one times the mass of the person, one times the radius squared and g is the constant that doesn't change. But when we compare it to the new planet, G is the constant so that doesn't change. If the mass of this planet is half the mass of earth, we have one half big M, little m doesn't change and then twice the radius, that means we have two times the radius but don't forget that square is distributed so that's 2 r squared which when we break it open is two squared times r squared.

So what does that give us? Let's cancel out all the constants or the values that change the same, the little m is dropped out, big M is dropped out, G drops out, the r squared drops out, and we're left with the proportionality of one times the force of gravity compared to one half divided by two to the second which is four. One half divided by four, we can multiply this as fraction four over one and when we want to divide we have to multiply by the reciprocal which means we have one half times one over four which means the new force on this planet is going to be one eight the force on planet earth. And the answer is one eight F (1/8F).

But you don't even have to go through every step, you can simply look at it as a proportionality and ask yourself, how does mass relate to force? Well because they're directly proportional if the mass of the planet goes down, the force will go down. If the radius is inversely proportional to force, when the radius goes up, force will go down. So if the radius goes up by a factor of two, we get two squared which is four and that means the force will be one quarter compare to the initial force. But in other way that you can use this is tying this equation to another force equation, for example, you may be ask to find the centripetal force on a satellite and you simply set, centripetal force equals gravitational force and solve for the unknown constants.

Be sure to join me in the next video where we look at the force of tension and you can find that video alongwith my entire Forces series, practice quiz and cheat sheet by visiting my website leah4sci.com/forces.

Are you stuck on a specific MCAT topic? I offer Private Online Tutoring where I focus on your needs to strengthen your individual weaknesses. Tutoring details can be found using the link below or by visiting my website leah4sci.com/MCATTutor.

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**[End Transcript]**

**Click here to Catch my entire video series on MCAT Physics Forces**

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