Acceleration given Earth's radius

AI Thread Summary
To calculate the acceleration of a Trojan Badger launched from a catapult at 8.5x10^8 m above Earth's surface, the gravitational acceleration formula is used, which incorporates both the distance from the Earth's center and the Earth's radius. The correct approach involves adding the Earth's radius (6.4x10^6 m) to the altitude (8.5x10^8 m) to find the total distance from the Earth's center. The gravitational acceleration can then be calculated using the formula g = (GM)/(r^2), where G is the gravitational constant and M is the mass of the Earth. The calculations confirm that the acceleration at that altitude is approximately 5.47 x 10^-4 m/s². This demonstrates the importance of considering the total distance when calculating gravitational effects at significant altitudes.
joel amos
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Given that the Earth has a radius of 6.4x10^6 m, what is the acceleration of a Trojan Badger launched from a catapult when it is 8.5x10^8 m above the surface of the Earth?

I'm not sure how to go about this question. All help is appreciated: formulas, explanations, answers, etc.
 
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joel amos said:
Given that the Earth has a radius of 6.4x10^6 m, what is the acceleration of a Trojan Badger launched from a catapult when it is 8.5x10^8 m above the surface of the Earth?

I'm not sure how to go about this question. All help is appreciated: formulas, explanations, answers, etc.

I believe it's just a gravity type question. Remember that a projectile's horizontal motion is at a constant velocity, neglecting air resistance. And the vertical motion is influenced by the acceleration of gravity.

What is the graviational equation that relates the attractive force between two masses and the separation distance? You should be able to use that to figure out the gravitational acceleration at that altitude. Please show your work.
 
The formula I know of is Fg = (G x m1 x m2)/ r^2 . However, I don't have the mass of the badger.
 
Here's what I've tried.
acceleration of gravity = [Radius of earth/Distance between objects]^2 x 9.8m/s/s

g = [6.4 x 10^6 m / 8.5 x 10^8 m]^2 x 9.8m/s/s
g = 5.56 x 10^-4 m/s/s

Is this correct?
 
joel amos said:
Here's what I've tried.
acceleration of gravity = [Radius of earth/Distance between objects]^2 x 9.8m/s/s
Distance between objects? What does that mean in this context?
 
haruspex said:
Distance between objects? What does that mean in this context?
Distance between the Trojan Badger and Earth. In other words, distance from earth.
 
joel amos said:
Distance between the Trojan Badger and Earth. In other words, distance from earth.
Think again. What answer would that give if it were 1mm from the surface of the earth?
 
Yeah, yeah, 4 x 10 ^ 20 m/s/s
I'm stumped. How would you solve this?
 
joel amos said:
The formula I know of is Fg = (G x m1 x m2)/ r^2 . However, I don't have the mass of the badger.

You don't need the badger's mass. What is the equation relating force to mass and acceleration?

And for the equation above, treat both masses as point masses for the purposes of calculating distance...
 
  • #10
F = ma

But, I don't know why I'd need to use the above equation to calculate distance...I already have distance.
 
  • #11
joel amos said:
F = ma

But, I don't know why I'd need to use the above equation to calculate distance...I already have distance.

That equation is why you don't need the mass of the badger. Combine the two equations, and be careful to think about what separation distance to use...
 
  • #12
So it'll be...
ma = (G*m*m)/r^s simplified to a = (G*m)/r^2 ?
where...
a = acceleration of gravity
G = gravitational constant
m = mass of earth
r = distance from Earth (8.5x10^8 m)
Is this correct?

And if the above is correct, then why would've my teacher included the radius of the Earth into the problem?
 
Last edited:
  • #13
joel amos said:
So it'll be...
ma = (G*m*m)/r^s simplified to a = (G*m)/r^2 ?
where...
a = acceleration of gravity
G = gravitational constant
m = mass of earth
r = distance from Earth (8.5x10^8 m)
Is this correct?

And if the above is correct, then why would've my teacher included the radius of the Earth into the problem?

You are on the right track. And the radius of the Earth was in your OP problem statement...
 
  • #14
And for the distance of separation... think centers of mass...
 
  • #15
Now I understand why the Earth's radius was in the original problem; it's because (as you hinted) the real distance is 8.5x10^8 m plus the radius of the Earth which is 6.4x10^6 m.
 
  • #16
joel amos said:
r = distance from Earth (8.5x10^8 m)
Distance from which bit of earth, exactly? Go back and think about why my 1mm example gave a silly answer.
 
  • #17
Is it this:
acceleration of gravity = [Radius of earth/(Distance from surface + Radius of earth)]^2 x 9.8m/s/s
 
  • #18
joel amos said:
Is it this:
acceleration of gravity = [Radius of earth/(Distance from surface + Radius of earth)]^2 x 9.8m/s/s
Yes.
 
  • #19
g = [Radius of earth/(Distance from surface + Radius of earth)]^2 x 9.8m/s/s
g = (GM)/(Distance from surface + Radius of earth)^2

So from this thread, I can conclude that both of the above formulas can be used to get an answer. Thankfully, they both yield 5.47 x 10^-4 m/s/s, confirming this to be the answer.

Thanks to everyone for the help.
 
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