Question

If mass increases, then the acceleration due to gravity....

Answer 1

It will stay the same in a vacuum where air resistance is not a factor. The acceleration due to gravity is always 9.8 m/s

Answer 2

If this is beginning physics, and you're learning about acceleration due to gravity, then acceleration at Earth's surface will be about \(-9.8[m/s^2]\) like @Claflamme3 said!

In reality, more massive objects will have slightly greater accelerations from gravity. So slightly, that it's not noticeable, and it's considered negligible in most physics situations.

Answer 3

I can show you the accepted math, where the larger mass means a greater acceleration, if you want.

Answer 4

Can you please? :) Thanks

Answer 5

It's for a lab, basically were dropping two masses, and we need to prove some sort of hypothesis

Answer 6

What did you say for your hypothesis?

Answer 7

If mass increases, then the acceleration due to gravity will remain constant because a freely falling object is solely based on gravity independent from the mass. Regardless of the mass, the acceleration due to gravity will remain the same

Answer 8

That is correct. Did you test your hypothesis yet? you will find that it is correct if you do

Answer 9

I did test it out, its just that that one was 7 and another was 10, but I think that's just due to systematic errors?

Answer 10

As @theEric said, in reality the larger mass wil fall a bit quicker... it may be because of systematic errors. How far are you dropping the objects?

Answer 11

A meter

Answer 12

When you said one was 10 and one was 7, are those seconds? If so then there probably was an error

Answer 13

I calculated the acceleration for both graphs, to find the 200g mass having 7.42/m^2 and the 500g mass having an acceleration of 10.3/m^2

Answer 14

You'll find it is highly certain! It will be as accurate as your measurements :)

The force due to gravity is given in this equation. The only variable you might not know is the gravitational constant, \(6.67384 \times 10^{-11} [m^3 /\ (kg\ s^{2})]\). The \(r\) is the distance between them, pretty much.
\[F_{gravity}=G\ \frac{m_{Earth}\ m_{object}}{r^2}\]

Answer 15

Ah that makes more sense. Like @theEric said, in reality the larger mass will have the faster acceleration. Maybe recalculate just to be sure

Answer 16

According to Newton's universal gravitation, the force an object r distance away from the Earth's center is approximately:
\[F_g=G \frac{ m_{earth} * m_{object} }{ r^2 }\]

We also know that:
\[F_g=m_{object} * a_{object}\]

Solving for acceleration, we get:
\[a_{object} = F_g / m_{object} = G \frac{ m_{earth} * m_{object} }{ r^2 } / m\]

Simplifying, we get:
\[a_{object} = G \frac{ m_{earth}}{ r^2 }\]

As you can see, the mass of the object is not part of the solution. So, the acceleration (caused by gravity) of an object is not dependent on the mass' mass.

Answer 17

I see the difference in accelerations. It is important, as @Claflamme3 said earlier, that air resistance is not a factor. The other important thing is, your measurement methods and tool have to be accurate!

Answer 18

Measurement tool- Ticker Tape

Answer 19

@queelius is correct, the acceleration will be constant! I was mistaken! The force will be different, but the acceleration is the same!

Answer 20

Thanks :) So, in this experiment I will mention what queelius said in my conclusion and explanation. I'll just say that the difference in accelerations is due to systematic errors

Answer 21

Therefore, it should come out to be 9.8 m/s. That is a good thing to write in your conclusion as well.

Answer 22

Now add your data, and what you found. Then state what could have gone wrong and how to improve on it.

Answer 23

OMG THANKS ALL OF YOU GUYS :) So I just state the acceleration that came out of my data, and the possible problems and solutions

Answer 24

No problem! Thats it.. good luck, i hope you get an A!

Answer 25

@Claflamme3 has the right idea about what to do with your data, just discussing it. Or you can try again and again and have all the data together.

Also, as @queelius pointed out, the field strength depends on the distance between the object and the Earth's center of gravity. So the acceleration will not always be \(9.8[m/s^2]\). It will be that at the Earth's surface.

Answer 26

Okk thanks, I'll mention all of this :)

Answer 27

Cool!

Answer 28

Well...OF COURSE the accelaration doesn't depend on the falling mass itself but it's just simply (G)*(Mearth)/(R)...
however, when you also think of air resistance, there will be a time when the velocity becomes steady no matter what accelaration... and the velocity will then depend on the mass, yes. because when you think of air resistance, ma=mg-kv and the steady velocity would be mg/k. So in reality and in great scale, the heavier it is, the quicker it will land on the ground. if you do not think of air resistance it would be ma=mg, making a=g which means that the accelaration doesn't differ or change.

Answer 29

@jh3power, note that the question states what is the acceleration due to gravity; not what is the acceleration.

Answer 30

Well...yes, of course... I was just trying to explain how experiments could turn out to be "different" from what we normally think...

Answer 31

@jh3power has a point, and it's relevant because there was a significant difference between the measured and real results. Air drag is a possibility for the difference in measurements from the expected, especially if air drag was not calculated into the expectation.

An example will be a hollow rubber-skin ball versus a more massive, same-sized kickball, with a thicker and heavier rubber skin. Air drag isn't immediately noticeable, but might be measurable, especially from a high enough drop distance.

I'm not sure of the math, but that point is valid.

There are also differences in measured values from expected values due to the measurement technique.

And, as hinted to before, some difference of measured from expected values can come from incorrect theory that was used to determine the expected value. Then the expected value might not match even well-measured results.