Gravitation: Calculating Accel. Due to Gravity at 1.43x10^8 m

[gjax21]

Homework Statement

1.A satellites are placed in a circular orbit that is 1.43 × 10^8 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance

Homework Equations

1.I used G*mE/r^2

The Attempt at a Solution

1. Got an answer of .0195m/s^2 I think I get this and have no idea why it is wrong.

 

[Doc Al]

What values did you use for G, mE, and r?

 

[RoyalCat]

Is the satellite $$1.43\cdot 10^8 m$$ above the surface of the earth, or $$1.43\cdot 108 m$$ above the surface of the earth?

Using the former, I got a result only slightly different from yours. Remember that the $$r$$ in that formula is the distance between the center of the Earth and the satellite, not from the surface of the Earth and the satellite.

 

[gjax21]

O thax I didn't realize it was saying above the surface. I meant 1.43x10^8. :)

using 6.674E-11 for G

Mof earth= 5.98E24

r=149380000

I get an answer of .0179, does that check out?

 

[gjax21]

Got it thax