# Can the gravity of an object be zero after a distance?

If you want to calculate the fall of an object on the earth, you have to consider the gravity of the earth. Should the gravity of the moon be considered? Or can it be taken as zero for the sake of calculation? When can this be wrong? That discussion is being done with a small test and some simple calculations.

Newton's law of gravitation tells us that the gravitational force F of an object (on another object) is inversely proportional to the square of R their distance. $F \proto \frac{1}{R^2}$

That is, as the distance increases, the force decreases, but never becomes zero. Nevertheless it is often written in the textbook that after a certain distance the value of this force decreases so much that it can be considered zero. But what exactly is that specific distance? Does the force of gravity have a definite value that can be taken to be zero if it is just as low? In this article we will try to explore these questions with a little experiment and some simple calculations.

Before I begin, let me ask the question again: How far can a gravitational force be zero? If anyone asks this question, he should first ask the counter question: Son, what are you doing for which the question of not considering the gravitational force is coming? What are you doing? This is important to know because the movement of the moon around the earth and the movement of its satellite Europa around Jupiter, these two figures must be taken into account.

Let's say I'm thinking about the world right now and doing a very simple calculation. While standing, I brought an orange to my head and released it. How long will it take for the lemon to fall to the ground? The gravity of the earth must be grasped to draw this figure. The question is, do you have to consider the gravity of the moon? Let's do that investigation.

The question of how long a lemon will take to fall to the ground can be answered in two ways. First of all, I can ask my father to take a video of me throwing my lemon on the camera of my smartphone. Then, using a video player app (such as VLC ), we can measure the time interval between the frame in which the video filer dropped the lemon and the frame in which the lemon touched the ground. See below for a detailed description of how I did this second step myself. (Let me say another small thing in this space. I assume we can measure this time interval up to one decimal place per second. Why this is important will be understood after a while.)

Secondly, I can try to calculate the time using Newton's law of gravity and the second law of motion. If you can do it properly, the two values ​​should be equal, right? Dad is in the kitchen right now, so we try to figure it out. The earth is pulling the lemon down by the force of gravity, so he falls down. Hold, the mass of the lemon m, the mass of the earth M, and the radius of the earth RThen according to Newton's law of gravitation, the amount of this force is: $F = \frac{G \cdot M \cdot m}{R^2}$

When the size of the force is known, we can calculate the acceleration of the lemon using Newton's second law of motion: $g= \frac{F}{m} = \frac{G \cdot M}{R^2}$

Interesting note: acceleration does not $m$depend on the mass of the lemon! That is to say, the value of this acceleration should be the same whether it is a lemon pill or a medicine pill or a stone pill. In this equation we find the universal gravitational constant $G$= 6.67 x 10 -11  m 3 kg -1 s 2 , $M$= 5.97 x 10 24 kg, and $R$= 6.37 x 10 6 m: $g = 9.8m/s^2$

Well, why don't we count the value of g here to two or three places after the decimal? The reason is that the number that I will $g$use to calculate the time, one digit after the decimal is enough to compare it with the time that my father will tell me. Because as I said before, I was able to measure up to a tenth of a second while taking time out of the video. For example, if the time in my calculation is 0.78 s, its 8 digits are meaningless in the context of my test. Because even if it is 6 or 7 instead of 8, I can't understand how accurate it is in the method of measuring time from video. So in that case we can write 0.78 s ≈ 0.8 s as the approximate value up to the first place after the decimal point.

Now we have to figure out how long it took to get out of this acceleration. Only then can it be seen whether it matches with the time measured from the hand-pen test. Suppose my height $h$= 5 feet = 1.5 m, and lemons take time to cover this distance $t$Then: $h = \frac{1}{2}gt^2\implies t = \sqrt\frac{2h}{g} \approx 0.6s$

When my father finished cooking, a video was taken of him throwing lemons and time was measured from him. Match!

[Dad, Mom, Aunt, PC, you can play this game by grabbing someone and getting a smartphone. Just remember that when you throw a lemon or tennis ball or medicine pill from a height, you have to calculate its value $h$ by replacing it with $t$ in the above equation.]

Let's come to the second part of the question. Not only the earth, but also the moon was pulling the lemon towards itself. Do we have to consider that attraction force in the figure? Let's calculate the acceleration ( $g\sp{\prime}$) of a lemon to attract the moon. The calculation of the previous chapter will also apply here. That is, $g\sp{\prime}= \frac{G \cdot M\sp{\prime}}{{R\sp{\prime}}^2}$

In this case the mass of the moon $M\sp{\prime} = 7.35 \times 10^{22} kg$and the distance of the moon from the earth $R\sp{\prime} = 3.84 \times 10^8 m$.

Without first $g\sp{\prime}$determining a perfect value, let's see if they $g\sp{\prime}$are $g$too small ( $g\sp{\prime} \ll g$), or too large ( $g\sp{\prime} \gg g$), or even close to each other ( $g\sp{\prime} \approx g$)? [Can you guess which of the three?] We can quickly calculate a "total" $G$ $M\sp{\prime}$and $R\sp{\prime}$use the power of the nearest 10 to replace its exact values ​​ : $G = 6.67 \times 10^{-11}m^3kg^{-1}s^2$ $\approx 10 \times 10^{-11} m^3 kg^{-1} s^2$ $= 10^{-10} m^3 kg^{-1} s^2$ $M\sp{\prime} = 7.35 \times 10^{22} kg \approx 10 \times 10^{22} kg = 10^{23} kg$ $R\sp{\prime} = 3.84 \times 10^8 kg \approx 1 \times 10^8 kg = 10^8 kg$

Let's put these values ​​in the equation: $g\sp{\prime} \approx \frac{10^{-10} \times 10^{23}}{(10^{8})^2} ms^{-2} = 0.001 ms^{-2}$

So it turns out that in the context of our little experiment, the moon is so far away from the earth that its gravity can be considered zero. But keep in mind that this is a contextual truth. If I had a device in my school that could measure t to the fourth place of a decimal, then the gravity of the moon could not be considered zero! Because $0.001m/s^{-2}$the difference will be caught then.

So it depends on which object's gravitational force I will hold and which I will assume to be zero, depending on which test I am comparing the results with and how precisely that test is going to be performed. If there is a difference in the answer to gravity of an object that cannot be tested by my instrument, then there is no benefit in gravity of that object, perhaps it can be called zero. For example, if the stopwatch in my figure can measure up to a decimal number, then the gravity of the moon can be said to be zero in the figure of throwing a lemon.

So to say, "After a certain distance, the gravity of an object becomes zero", does not make sense. It may be said, “It is true that an object has gravity after a certain distance, but the difference in the answer to the sum is small. If I can't measure the slightest difference in my instrument, then there is no benefit in gravity of the object. ” Sometimes things are said simply to facilitate discussion. The same thing applies with “gravity becomes zero”. But to put it simply, a misconception can remain in the head, which can lead to problems in certain cases. For example, if there is a fine time measuring instrument, if there is no gravity of the moon, then there will be a problem.

 This type of quick calculation is a great tool for scientific research. Scientists call this back-of-envalope calculation.

How to measure time from video file

I used a tennis ball instead of a lemon to do the test. You can use any unbreakable object.

1.  I asked my friend to take a video of me throwing the ball on the camera of his smartphone. I also said that in the video both my hand with the ball and the ground where the ball will fall can be seen. Watch the video file above. The ball has been dropped from a height of 5 feet here. I measured this height with tape before taking the video.
2. I transferred the video file from the smartphone to my computer.
3. I opened the video on   VLC player on the computer .
4. One by one frames of the video opened in VLC can be advanced by pressing the E button on the keyboard . So I saw the frame forward, there are 30 frames in the first one second. That means the average time interval between two frames is 1/30 second.
5. When the ball got out of my hand, I started by pressing the E button and moving the frame one by one. The ball touched the ground after advancing 17 frames. Then the ball took time to fall feet t = 17/30 seconds = 0.6 seconds.

Another thing: once the test is done, there may be a slight error in size for various reasons. In order to cut it, the test should be done several times and all the standards should be averaged. This applies not only to this test, but to any test in science.

1. 