submitted 8 months ago by GardenVarietyAnxiety@lemmy.world to c/physics@mander.xyz

4 comments fedilink hide all child comments

Is it possible to determine the percentage of the gravitational force at a specified distance using only the geometry of the planet?

Example: The ISS at ~420km altitude "weighs" about 90% of what it would on the Earth's surface.

Is there an equation using only geometrical values that would give you this info?

you are viewing a single comment's thread
view the rest of the comments

[-] nooneescapesthelaw@mander.xyz 3 points 8 months ago* (last edited 8 months ago)

~~Mass~~ weight of object on earth = g(mass of object * mass of earth)/(earth radius)^2

Weight of object in orbit= g(mass of object * mass of earth)/(earth radius + distance from earth)^2

Dividing the two equations gives us:

(1/(earth radius + distance from earth)^2)
 ---------------------------------------
(1/earth radius)^2

Which yields:

(Earth radius)^2
-----------------------------------------
(earth radius + distance from earth)^2

Multiplying the fraction above by 100 will give a percentage.

To check if the equation makes sense imagine if the object was very far from earth. Then the weight (earths gravity) would be small and thus the percentage would be small. This checks with our equation

We can try it for the ISS and see we get:

((6378)^2) ÷(408+6378)^2 = .88

Or 88%

[-] GardenVarietyAnxiety@lemmy.world 1 points 8 months ago

This was helpful... I appreciate you taking the time to write it all out~

this post was submitted on 23 Jan 2024

7 points (81.8% liked)

Physics

1316 readers

3 users here now

founded 2 years ago

MODERATORS

Sal@mander.xyz