Calculating Acceleration Due to Earth's Gravity at Moon's Orbital Radius

[MozAngeles]

Homework Statement

PHYSICS Gravity exercise!?
What is the acceleration due to Earth's gravity at a distance from the center of the Earth equal to the What is the acceleration due to Earth's gravity at a distance from the center of the Earth equal to the orbital radius of the Moon?
a = --------------m/s²
where to start? use Gm₁m₂/r²? and a=v²/r?

Homework Equations

The Attempt at a Solution

no idea where to start... iguess I am not sure exactly what the question is really asking...

 

[cepheid]

Homework Statement

PHYSICS Gravity exercise!?

LOL! I can't figure out why you seem so surprised by this...

What is the acceleration due to Earth's gravity at a distance from the center of the Earth equal to the What is the acceleration due to Earth's gravity at a distance from the center of the Earth equal to the orbital radius of the Moon?
a = --------------m/s²

I think you didn't finish typing what the first radius was supposed to be...

where to start? use Gm₁m₂/r²? and a=v²/r?

Newton's law of gravitation is a great place to start. In fact, it is exactly what you need. The second formula is not relevant, since it is the formula for centripetal acceleration, which only occurs during circular motion.

The Attempt at a Solution

no idea where to start... iguess I am not sure exactly what the question is really asking...

What is the acceleration due to gravity? In other words, what acceleration does the object experience as a result of the force of gravitational attraction between that object and Earth? You know from Newton's second law that the acceleration it experiences will be given by the net force applied to it, divided by its mass. In this case, the only force on it is Earth's gravity. You have an expression for this gravitational force on it. So all you have to do is divide that force by the mass of the object to get the acceleration. (Actually, the expression you have is a generic expression for the mutual gravitational force between any two masses m1 and m2. In this case, one of those masses is the object in question, and the other one is Earth). So, the acceleration, in the end, will depend only upon three parameters: a fundamental constant G, the mass of the Earth M, and the distance r. You just need to compute this twice, once for each of the two distances.

 

[MozAngeles]

so i typed it wrong the first time this is the full question. "What is the acceleration due to Earth's gravity at a distance from the center of the Earth equal to the orbital radius of the Moon?"
and i understand that you need to use Gm₁m₂/r², but I'm wondering if the radius i would use is the distance between the moon and the earth. and for the mass 1 use the mass of the Earth and mass two the mass of the moon. Then set this equation equal to m₂g. So then g will be the gravitational acceleration...

 

[cepheid]

so i typed it wrong the first time this is the full question. "What is the acceleration due to Earth's gravity at a distance from the center of the Earth equal to the orbital radius of the Moon?"
and i understand that you need to use Gm₁m₂/r², but I'm wondering if the radius i would use is the distance between the moon and the earth.

Well yeah, I mean, what is the radius of the moon's orbit? That is exactly the distance that the problem asks you to use.

and for the mass 1 use the mass of the Earth and mass two the mass of the moon.

Yeah, like I said before, one of the two masses is Earth. It's irrelevant what mass you use for the second one, since m2 will cancel from both sides of the equation. In other words, ANY mass at that distance from Earth will experience the same acceleration.

Then set this equation equal to m₂g. So then g will be the gravitational acceleration...

Yes. The nice thing about this exercise is that it shows you where g comes from...how it can be expressed in terms of more fundamental constants like G and the mass of the Earth and (in the case of Earth surface gravity) the radius of the earth.

 

[MozAngeles]

Sorry to not be getting this too well, its probably really easy to most people. But to make sure..
GM(earth)m/R(moon's orbital radius)=mg
then m cancels
g=G*5.97E24/3.84E8 = 1.04E6 m/s
but this is wrong according to the back of the book. So where am I going wrong?

 

[HallsofIvy]

Sorry to not be getting this too well, its probably really easy to most people. But to make sure..
GM(earth)m/R(moon's orbital radius)=mg
then m cancels
g=G*5.97E24/3.84E8 = 1.04E6 m/s
but this is wrong according to the back of the book. So where am I going wrong?

Force due to gravity = [tex]\frac{GMm}{r^2}[/tex]
You seem to have forgotten to square r.

 

[MozAngeles]

Thank you sooooo much , it is quite easy... I just got consfued