Calculating the Hill sphere of the International Space Station

Click For Summary
The discussion focuses on calculating the Hill sphere of the International Space Station (ISS), which has a mass of 400,000 kg and orbits 408 km above Earth's surface. Participants clarify that the 109 m measurement refers to the physical diameter of the ISS, not the Hill sphere. The height of 408 km is the distance above Earth's surface, not the total distance to the center of Earth. Visualizing the problem through sketches is recommended to aid understanding. The conversation concludes that while a positive Hill sphere calculation could suggest the possibility of another object orbiting the ISS, it must not intersect with the station itself.
fema98
Messages
3
Reaction score
0

Homework Statement

[/B]
The International Space Station (ISS) has a mass of 400,000 kg and orbits 408 km
above the Earth’s surface. The ISS is 109 m across.

Homework Equations

: [/B]
R=a(semimajoraxis) cubedroot(m2/3M1)

The Attempt at a Solution

: [/B]
ive tried multiple ways with multiple equations, I just can't figure out what the variable are sipposed to be I am assuming that M1 is 400,000Kg, and M2 isn't given my main problem is what do 408Km and 109M represent, is the 109 M the Hill sphere, so would the euation work out to be 109=204 cubedroot (m/3*400,000kg)? or is this just a conceptual question and I am way overthinking this?

PS. if your wondering where the 204 came from i assumed the 408 distance from Earth to the ISS was the major axis so semimajor was 204.
 
Last edited by a moderator:
Physics news on Phys.org
fema98 said:
and M2 isn't given my main problem
What does the ISS orbit? You can look up the mass of this object.
fema98 said:
is the 109 M the Hill sphere
No, it is the physical diameter of the space station.

408 km is the height of the space station above the surface of Earth. That is written in the problem statement, I don't see what could be unclear here.
fema98 said:
PS. if your wondering where the 204 came from i assumed the 408 distance from Earth to the ISS was the major axis so semimajor was 204.
(a) work with units, then you would have spotted one error here already
(b) 408 km is the height of the ISS above the surface of Earth (that is literally given in the problem statement), it is not the distance to the center of Earth.

Did you draw a sketch?
 
Firstly thanks for the clarification. And no i didn't draw it, never been able to visualize math/physics problems. Also i would presume that aslong as i get a positive number than that would indicate that theoretically there could be another station that could orbit the ISS aslong as it was within that Hill sphere?
 
fema98 said:
And no i didn't draw it, never been able to visualize math/physics problems.
That makes it difficult to solve problems. Drawing a sketch will help to visualize it.
fema98 said:
Also i would presume that aslong as i get a positive number than that would indicate that theoretically there could be another station that could orbit the ISS aslong as it was within that Hill sphere?
Not if the orbit would have to physically go through the station.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
2
Views
4K
Replies
10
Views
7K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K