Using ratios to obtain the weight of a person at a specific distance

Click For Summary
SUMMARY

The discussion focuses on calculating the weight of a person at a height of 128,000 km above Earth's surface, given their weight of 980 N on the surface. The key equation used is the inverse square law, expressed as Fg is proportional to 1/d². The solution reveals that at 128,000 km, which is 20 times the Earth's radius, the weight reduces to 0.072 N. Participants emphasize the importance of understanding ratios and proportionality constants in gravitational force calculations.

PREREQUISITES
  • Understanding of gravitational force and its relationship to distance
  • Familiarity with the inverse square law in physics
  • Knowledge of ratios and proportional relationships
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the inverse square law in gravitational physics
  • Learn how to calculate gravitational force at various distances from a celestial body
  • Explore the concept of proportionality constants in physics
  • Practice solving problems involving weight changes at different altitudes
USEFUL FOR

Students studying physics, educators teaching gravitational concepts, and anyone interested in understanding weight variations due to altitude changes.

needingtoknow
Messages
160
Reaction score
0

Homework Statement



Assuming the person weighs 980 N on the surface of the Earth, how would you use ratios to obtain the weight of a person at 128 000 km above the surface of the Earth.


Homework Equations



Fg is proportional to 1/d2


The Attempt at a Solution



The answer in the key says 0.072 N. I didn't know exactly where to start but by working backwards from the answer I have determined that it is 20 times the distance from the centre of Earth. I know how to solve these questions using ratios if I am given the times the distance from the centre of Earth.

For example for a previous question: Three times the distance from the centre of Earth, here is how I solve it.

Fg is proportional to 1/d^2

1x3 = 3 therefore 980 = 1/3^2

980 / 9 = 109 N

This method works everytime when I am given the times the distance from the centre of Earth. But when I am given a specific distance I do not know how to get the answer. Can someone please help? Thanks!
 
Physics news on Phys.org
Since F\propto\frac{1}{d^2}, You can write F = k\frac{1}{d^2} where k is the constant of proportionality.

So if F_1\propto\frac{1}{d_1^2} and F_2\propto\frac{1}{d_2^2}

then \frac{F_1}{F_2}=\frac{\frac{1}{d_1^2}}{\frac{1}{d_2^2}}.

The constants of proportionality cancel.
 
The radial distance from the center of the Earth to the surface is ? What is the ratio of 128000 km to this distance?
 

Similar threads

Replies
6
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
5
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
17
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
5
Views
2K