physics

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!

1. "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.

posted by Marth
2. 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.

posted by drwls

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