Physics problem... Stumped (1 Viewer)

[SWJJ]

http://www.daviddarling.info/encyclopedia/A/accgrav.html

The acceleration that an object experiences because of gravity when it falls freely close to the surface of a massive body, such as a planet. Also known as the acceleration of free fall, its value can be calculated from the formula


g = GM / (R + h) 2

M should be the mass of the earth plus the mass of the sat, I will assume the mass of the sat is inconcesquential compared to the mass of the earth and work the proble set up like that..

 

[Saints#34]

http://hyperphysics.phy-astr.gsu.edu/hbase/orbv.html


g(orbit)=g(earth)[R(earth)/h]^2


edit: I haven't taken physics in a while so I don't remember all the formulas, but I seem to remember both the mass of the satellite and the mass of the earth cancelling out in problems like this.

 

[SWJJ]

^^^ he does not know the mass only the distance.

For some reason this strikes me as a quick question if the satellite is just orbiting the earth with no acceleration other than the pull of gravity wouldn't the answer simple be 9.82m/s^2.

Or is it different since it is in space and not here on earth?

Just a thought to look at

T

It should be less because of the difference in distance

I calculated 4.63 m/s^2 and am going with that its logical.

9.8 m/s on the surface (distace of 6.38 x 10 ^6 - earths radius) and a hieght of

2.90 × 10 ^6

slightly less than the same hieght as the raduis of the earth so it makes sense g should be slightly more than half g on the surface...

 

[Saints#34]

Thats what the website above gave when I plugged in 2.9x10^6m, so I think you are safe.

 

[efil4stnias]

A satellites are placed in a circular orbit that is 2.90 × 10 ^6 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?


not sure how to set this up...

Whew...glad i skipped Physics from High School thru college. I have no clue but would answer pi. 2.13?

 

[buzd]

M should be the mass of the earth plus the mass of the sat, I will assume the mass of the sat is inconcesquential compared to the mass of the earth and work the proble set up like that..

If you assume the mass of the sat to be inconsequential, aren't you assuming it to be zero? Therefore there wouldn't be any gravity.

Q.E.D.

 

[johnnyd]

try here:
http://hypertextbook.com/physics/mechanics/gravitation-extended/problems.shtml

 

[ra]

acceleration is not just a change in velocity, but a change in direction as well.

No it isn't.
Acceleration is by definition the change in velocity.
Velocity is a vector quantity.
A change in direction will change your velocity which will change your acceleration.
Satellites aren't changing direction normal to the center of the object they are circling.

The force that pulls the sat down towards the earth, instead of it flying straight into space, is the acceleration.

....

If the force keeping the object in a set orbit around the Earth due to centripetal force (the force pulling the object due to the gravity of the Earth) is equal to the force the object feels normal to the centripetal due to its rotational velocity, the acceleration is equal to zero.

 

[buzd]

Acceleration is by definition the change in velocity.

However, velocity is defined as rate of movement in a given direction. Without the direction, it's just speed.

Therefore, change direction implies change velocity.

 

[ra]

However, velocity is defined as rate of movement in a given direction. Without the direction, it's just speed.

Therefore, change direction implies change velocity.

But there is also angular velocity.
The angular velocity remains constant with an orbital.

 

[Jeff Miller]

acceleration is not just a change in velocity, but a change in direction as well.

The force that pulls the sat down towards the earth, instead of it flying straight into space, is the acceleration.

I dont want the answer jonesy, just how to set it up, I dont have the mass of the sat so I cant calculate it that way... not sure where to go..

well i'll be snickered....

In physics, acceleration is defined as the rate of change of velocity, or as the second derivative of position (with respect to time). It is then a vector quantity with dimension length/time². In SI units, acceleration is measured in meters/second² (m·s-2). The term "acceleration" generally refers to the change in instantaneous velocity.

In common speech, the term acceleration is only used for an increase in speed; a decrease in speed is called deceleration. In physics, any increase or decrease in speed is referred to as acceleration and similarly, motion in a circle at constant speed is also an acceleration, since the direction component of the velocity is changing. See also Newton's Laws of Motion.

 

[ra]

A satellites are placed in a circular orbit that is 2.90 × 10 ^6 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?


not sure how to set this up...

If all you need to find is the force of gravity on the object, it's

Fg=G(m1m2/r^2)

Remember to add the radius of the Earth to 2.90 × 10 ^6 m to get the value for r.

 

[Det. Brees]

actually i think as the orbit decays it begins to fall to the earth shortening its orbit length and causing it to increase in speed....
just a thought. no facts to back it up...

 

[wbbigtymer]

No it isn't.
Acceleration is by definition the change in velocity.
Velocity is a vector quantity.
A change in direction will change your velocity which will change your acceleration.
Satellites aren't changing direction normal to the center of the object they are circling.



If the force keeping the object in a set orbit around the Earth due to centripetal force (the force pulling the object due to the gravity of the Earth) is equal to the force the object feels normal to the centripetal due to its rotational velocity, the acceleration is equal to zero.

You have the normal component and the tangential component. There is no tangential acceleration since the rotational velocity is constant, but there is a normal acceleration (towards earth), which would be defined as V^2/R.