Hi i'm doing that lab were you attach a weight to a tape which is also through a timer that puts marks on the tape at a given time interval

ok I'm asked why is gravity different than on Earth?

I don't rember the physics answer behind it I remeber

Fg = G((m1 m1)/r^2)

to find the force of gravity but I don't really remeber like the answer...

like what is different in that equation for the Moon than the Earth...

Thanks!!!

ok also I'm asked for the lab

Was there a difference between the accelerations for the different masses? Explain why?

I this occurs because of a drag force acting on the weight. A larger weight is going to have a larger reference area right so the drag force would be larger which would off set newtons second law of motion upseting the net force and changing the acceleration

Is that correct?

ok for the lab

I'm also asked Would you expect your calculated value for acceleration to be higher than, equal to, or less than, the accepted value for acceleration due to gravity (9.80 m/s^2) why?

Ok for this one I remeber some how using this equation...

Fg = G((m1 m1)/r^2)

to find gravity and I could use that equation to find the gravity and then use that for my reason why

the only thing is I don't remeber how to find gravity given your cordinates

I remeber it was very simple to find the force of gravity at the equator and how it's like 3 thousands less and how people argued that it would make a difference and world records would be broken at the olympics or something like that...

THANKS!

"what is different in that equation for the Moon than the Earth?"

The mass of the body (Earth or moon) as well as the distance (radius, if one is on the surface) is different.

However, when using that equation to calculate gravitational forces between several lab objects, unless the objects are extremely massive the gravitational forces between them will be slight, almost negligent compared to the force of weight.

Drag will decrease the measured acceleration rate below the nominal 9.81 m/s^2 value.

The actual value of computed g, assuming no drag, also varies somewhat over the earth's surface due to the earth's spin (which causes a latitude-dependent centrifugal force to get subtracted), lack of a perfect spherical shape of the earth, and the proximity of very high mountain ranges.

To answer your questions, let's start with the first one about the difference in gravity between the Moon and Earth.

The equation you mentioned, Fg = G((m1 m2)/r^2), represents the force of gravity between two objects. In this equation, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

In the case of the Moon and Earth, the main difference lies in the mass of the two objects. The Moon has much less mass compared to Earth. Since the force of gravity is directly proportional to the masses of the objects involved, the force of gravity on the Moon is much weaker than on Earth. Therefore, the value of 'm2' (mass of the Moon) would be smaller than 'm1' (mass of the Earth) in the equation you provided, leading to a smaller force of gravity on the Moon.

Now let's move on to your second question about the differences in acceleration for different masses in your lab.

Your explanation is partially correct. In the lab, if there is a drag force acting on the weight attached to the tape, then the larger weight would experience a greater drag force compared to the smaller weight. This additional drag force on the larger weight would indeed affect the net force acting on it and therefore change the acceleration.

However, it's important to note that the difference in acceleration is not solely due to the drag force. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. So, for the same net force, a smaller mass will result in a larger acceleration, while a larger mass will result in a smaller acceleration. Therefore, in your lab, you should expect the smaller masses to have higher accelerations compared to the larger masses, assuming the net force is constant.

Finally, for your third question about the calculated value of acceleration compared to the accepted value for acceleration due to gravity (9.8 m/s^2), let me clarify the relationship between gravity and the formula you mentioned earlier.

The formula Fg = G((m1 m2)/r^2) calculates the force of gravity between two objects. To calculate the acceleration due to gravity (g), you need to consider the force of gravity acting on a single object. This can be done using Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration.

So, to find the value of gravity (g), you can rearrange the formula as follows:

Fg = m * g (where m is the mass of the object)

G((m1 m2)/r^2) = m * g

g = G * ((m1 m2)/r^2) / m1

In this equation, you would substitute the values of G, m1, and r with the appropriate values for the location you are interested in (e.g., Earth's equator). From this calculation, you can compare the calculated value of g to the accepted value of 9.8 m/s^2. Depending on the location, the calculated value may be slightly higher or lower than the accepted value due to variations in gravitational acceleration at different locations on Earth.

I hope this explanation helps clarify your questions. Let me know if you need further assistance!