# Question Video: Calculating the Acceleration Due to Gravity of an Object Physics • 9th Grade

The diagram shows two objects, A and B. What is the initial acceleration of object B due to its gravitational interaction with object A?

06:10

### Video Transcript

The diagram shows two objects, A and B. What is the initial acceleration of object B due to its gravitational interaction with object A?

As we can see, our diagram contains two objects: object A in the upper left and object B in the lower right. There will be a gravitational force acting between these two objects pulling object A towards object B and with an equal magnitude and opposite direction pulling object B towards object A. This force will cause each of the objects to accelerate towards each other. And the question asks us to work out what the acceleration of object B will be.

Let’s start by looking at the force acting between these two objects, which we will call capital 𝐹. The gravitational force between two objects can be worked out using Newton’s law of gravitation. And this law tells us that the force acting between two objects is equal to big 𝐺, the gravitational constant, multiplied by the product of the two masses of the objects divided by the distance between the objects’ centers of mass squared.

In our case, we will say that 𝑚 one refers to the mass of object A. So, 𝑚 one is equal to 1,800,000 kilograms. And we’ll say that 𝑚 two refers to the mass of object B. So, 𝑚 two is equal to 2,900,000 kilograms. We should also note that big 𝐺 has a defined value of 6.67 times 10 to the power of negative 11 meters cubed per kilogram second squared. So, we know the values of the masses and the gravitational constant. The only thing we don’t know at the moment is the distance between the objects. Luckily, we can use our diagram to work this out.

We can see that in our diagram, one square horizontally represents one kilometer in the 𝑥-direction, and the same vertically for the 𝑦-direction. So, if we count the number of squares in the 𝑥- and 𝑦-direction between the centers of mass of the objects, we can calculate the distance between the objects in the 𝑥- and 𝑦-directions, respectively. We’ll label this distance in the 𝑥-direction 𝑟 sub 𝑥 and in the 𝑦-direction 𝑟 sub 𝑦. So, in the 𝑥-direction, we have one, two, three, four, five, six squares. And because each of these represents one kilometer, 𝑟 sub 𝑥 is equal to six kilometers.

In the 𝑦-direction, we have one, two, three, four, five, six, seven, eight squares. So, we know that 𝑟 sub 𝑦 is equal to eight kilometers. We can use these to work out the distance between the objects’ centers of mass, which we have called 𝑟, by recognizing that 𝑟, 𝑟 sub 𝑥, and 𝑟 sub 𝑦 form a right-angled triangle. We can therefore use Pythagoras’s theorem to calculate 𝑟.

Pythagoras’s theorem states that the square of the hypotenuse of a right-angled triangle, in our case 𝑟 squared, is equal to the sum of the squares of the remaining two sides, in our case 𝑟 sub 𝑥 squared plus 𝑟 sub 𝑦 squared. Substituting our known values of 𝑟 sub 𝑥 and 𝑟 sub 𝑦 into this, we get 𝑟 squared is equal to six kilometers squared plus eight kilometers squared. Six kilometers all squared is equal to 36 kilometers squared, and eight kilometers all squared is equal to 64 kilometers squared. Then, 36 kilometers squared plus 64 kilometers squared is equal to 100 kilometers squared. And then, finally, taking the square root of this will give us a value for 𝑟. The square root of 100 kilometers squared is simply 10 kilometers. So, the distance between these two objects is 10 kilometers, and we’ll keep a note of this up here on the right.

So, we now have values for everything we need to calculate the force acting between the two objects. But how does this actually relate to the acceleration of object B? Well, we know from Newton’s second law that the force acting on object B is equal to the mass of object B multiplied by its acceleration. So, we can write that the acceleration of object B is equal to the force acting on it divided by its mass.

Substituting our expression for force given to us by Newton’s law of gravitation, the acceleration of object B is equal to big 𝐺 multiplied by 𝑚 one multiplied by 𝑚 two divided by 𝑟 squared all divided by 𝑚 two. And what we can see is that we have 𝑚 two in the numerator and the denominator of this fraction. So, they will cancel, leaving us with just big 𝐺 multiplied by 𝑚 one divided by 𝑟 squared. And what we can see here is that the acceleration of object B only depends on the mass of object A and the distance between them. Object B could be anything; it could be an ant or a person, or it just so happens in this question that it has a mass of 2,900,000 kilograms.

All that’s left for us to do now is to substitute our known values of 𝑚 one, big 𝐺, and 𝑟 into this equation. But first we must check our units. For all of our values, 𝑚 one, 𝑚 two, and big 𝐺, when there is a mass, we see that it is expressed in kilograms. Similarly, when time appears in big 𝐺, it is in seconds. However, when we look at the units of distance, we see that they’re inconsistent. Big 𝐺 has units of distance which are meters, whereas our distance 𝑟 is expressed in kilometers.

In order to continue, these must be consistent. So, we’ll convert 𝑟 to meters. We can recall that one kilometer is equal to 1,000 meters. So, 10 kilometers is equal to 10,000 meters. This means that any time distance appears in any of our values, it is in meters. So, we can continue and substitute these into our equation for acceleration.

The acceleration of object B is equal to 6.67 times 10 to the power of negative 11 meters cubed per kilogram second squared multiplied by 1,800,000 kilograms divided by 10,000 meters squared. Evaluating this gives us a value of 𝑎 of 1.20 times 10 to the power of negative 12 meters per second squared. And this is the acceleration of object B.

So, the initial acceleration of object B due to its gravitational interaction with object A is equal to 1.20 times 10 to the power of negative 12 meters per second squared.