How Does Distance from Earth's Center Affect Gravity?

Click For Summary

Homework Help Overview

The discussion revolves around the gravitational force experienced by an astronaut at different distances from the center of the Earth, specifically comparing the force at the surface to that at a distance of 2 meters from the center, expressed in multiples of Earth's radius.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the idea of using ratios of gravitational forces to simplify calculations. There is a suggestion to express the forces as a ratio to see if any terms cancel out. Some participants express confusion about the initial setup and the lack of mass information.

Discussion Status

Participants are actively engaging with the problem, discussing potential methods to approach it. Some guidance has been offered regarding the use of ratios, but there is no clear consensus on the understanding of the problem setup or the next steps.

Contextual Notes

The original poster notes a lack of mass information, which may be affecting their ability to proceed with the calculations. There is also a mention of needing clarification on the suggested approaches.

King Khan
Messages
3
Reaction score
0

Homework Statement


The force of gravity at Earth's surface on an astronaut if 634N. What is the force of gravity on the same person if the distance is 2m, in multiplies of Earth's radius, from the center of Earth?

Homework Equations


F=Gm1m2/r^2

The Attempt at a Solution


I really don't understand what to plug in where seeing as how they don't even give the mass, if someone could just please help me with this part I could do the rest.
 
Physics news on Phys.org
Why don't you write it as a ratio of two forces and see if anything cancels.

Fa/Fb

Where Fb is the force at the surface and Fa is the force at twice the radius.
 
DaleSpam said:
Why don't you write it as a ratio of two forces and see if anything cancels.

Fa/Fb

Where Fb is the force at the surface and Fa is the force at twice the radius.
if you could rephrase that, sorry but I really don't understand.
 
[tex]F_a=\frac{G \, m1_a \, m2_a}{r_a^2}[/tex]
[tex]F_b=\frac{G \, m1_b \, m2_b}{r_b^2}[/tex]
Write Fa/Fb, simplify and solve.
 

Similar threads

Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
Replies
2
Views
1K
Replies
3
Views
3K
Replies
10
Views
4K
  • · Replies 51 ·
2
Replies
51
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
8
Views
3K
Replies
18
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K