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

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Homework Help Overview

The problem involves calculating the acceleration due to gravity at a distance of 1.43 × 10^8 m above the Earth's surface, specifically in the context of satellite motion and gravitational forces.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the values used for gravitational constant, mass of the Earth, and the correct interpretation of the distance from the Earth's center versus the surface. There is an exploration of the calculations leading to different results.

Discussion Status

Some participants have provided alternative calculations and clarified the interpretation of the problem. There appears to be ongoing exploration of the correct values and their implications on the results, but no explicit consensus has been reached.

Contextual Notes

Participants are addressing potential misunderstandings regarding the distance measurement and the application of the gravitational formula. The original poster initially misinterpreted the distance as being from the surface rather than from the center of the Earth.

gjax21
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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.
 
Last edited:
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What values did you use for G, mE, and r?
 
Is the satellite [tex]1.43\cdot 10^8 m[/tex] above the surface of the earth, or [tex]1.43\cdot 108 m[/tex] above the surface of the earth?

Using the former, I got a result only slightly different from yours. Remember that the [tex]r[/tex] 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.
 
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?
 
Last edited:
Got it thax
 

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