Magnitude of acceleration due to gravity

In summary, the conversation discusses calculating the magnitude of acceleration due to gravity for a satellite placed in a circular orbit 1.20 × 106 m above the surface of the Earth. The relevant equations are W = G((ME m)/r2) and W = mg, and the solution involves substituting the values into the equation g' = (Re^2/r^2)*g, where r is the distance to the center of the Earth and g is the acceleration due to gravity. The final calculated value is 276.53 m/s^2.
  • #1
Shadow236
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Homework Statement


Satellites are placed in a circular orbit that is 1.20 × 106 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?



Homework Equations


My book is saying that the relevant equations are:

W = G((ME m)/r2)

W = mg


The Attempt at a Solution


After looking at this post, I tried the formula g = MG/r^2. (M being the mass of the Earth)

g = ((6.67 x 10^-11)(5.97 x 10^24))/((1.20 x 10^6)^2)

This^ yielded 276.53 m/s^2, which is apparently wrong.
 
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  • #2
g' = (Re^2/r^2)*g

r = Re + h where Re is the radius of the Earth and h is the height above it.

g = acceleration due to gravity on the earth

substitute the values given in the equation in that formula and you should get the answer for g'
 
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  • #3
As Ronaldo said, ##r## is the distance of the object to the CM of earth. That means, as Ronaldo stated: $$r=1,20*10^6+r_{earth}$$
 
  • #4
It looks like you guys are correct, because I got the answer correct and the book says the following...: http://twitpic.com/dx955g

Thank you!
 
  • #5


Your attempt at using the formula g = MG/r^2 was correct. However, the value you used for the radius (1.20 x 10^6 m) was incorrect. The radius in this case should be the distance from the center of the Earth to the satellite, which is 6.37 x 10^6 m (radius of the Earth + altitude of the satellite).

Correct calculation:

g = ((6.67 x 10^-11)(5.97 x 10^24))/((6.37 x 10^6 + 1.20 x 10^6)^2)

g = 9.81 m/s^2

The magnitude of the acceleration due to gravity at this distance is 9.81 m/s^2. This means that objects in this orbit will experience the same gravitational acceleration as objects on the surface of the Earth.
 

FAQ: Magnitude of acceleration due to gravity

1. What is the magnitude of acceleration due to gravity?

The magnitude of acceleration due to gravity is a constant value of 9.8 meters per second squared (m/s²) on Earth. This means that every second, an object will accelerate 9.8 m/s² towards the ground due to the force of gravity.

2. How is the magnitude of acceleration due to gravity calculated?

The magnitude of acceleration due to gravity can be calculated using the formula a = g, where "a" represents acceleration and "g" represents the gravitational constant of 9.8 m/s².

3. Does the magnitude of acceleration due to gravity change with altitude?

Yes, the magnitude of acceleration due to gravity does change with altitude. This is because the force of gravity is dependent on the distance between two objects, and as altitude increases, the distance between an object and the center of the Earth also increases. This results in a slightly lower magnitude of acceleration due to gravity.

4. How does the magnitude of acceleration due to gravity differ on other planets?

The magnitude of acceleration due to gravity differs on other planets due to differences in mass and radius. For example, the magnitude of acceleration due to gravity on Mars is 3.7 m/s², while on Jupiter it is 24.8 m/s².

5. Can the magnitude of acceleration due to gravity be altered?

The magnitude of acceleration due to gravity cannot be altered by humans. It is a fundamental force of nature that is constant and cannot be changed. However, it can be affected by factors such as altitude, mass, and distance from other objects.

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