A community for students.
Here's the question you clicked on:
 0 viewing
idku
 one year ago
Advanced PHYSICS question. SATELLINE in SPACE (((FORCES, ACCELERATIONS etc...)))
idku
 one year ago
Advanced PHYSICS question. SATELLINE in SPACE (((FORCES, ACCELERATIONS etc...)))

This Question is Closed

idku
 one year ago
Best ResponseYou've already chosen the best response.0Scientists want to place a 2700 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 2.3 times the radius of Mars above the surface of the planet. Here is some information that will help solve this problem: dw:1442544068910:dw \(\large \rm m_{mars} = 6.4191 \times 10^{23} kg \) \(\large \rm r_{mars} = 3.397 \times 10^6 m \) \(\large \rm G = 6.67428 \times 10^{11} {~}(Nm^2)/kg^2 \)

idku
 one year ago
Best ResponseYou've already chosen the best response.01) What is the force of attraction between Mars and the satellite?

idku
 one year ago
Best ResponseYou've already chosen the best response.0((I dissapear from the site, due to this site's faultiness when it comes to connecting. Apologize ))

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0you may use newton's law of gravity : \[F = G \dfrac{m_1m_2}{r^2}\]

idku
 one year ago
Best ResponseYou've already chosen the best response.0G × (2700kg × 6.4191×10\(^{23}\)kg) / (3.397 × 10\(^{6}\)m) and I will plug my G of course...

idku
 one year ago
Best ResponseYou've already chosen the best response.06.67428×10^(11)• (2700 × 6.4191×10^(23)) / (3.397 × 10^(6)) this is what I plugged into wolfram, and got 3.41

idku
 one year ago
Best ResponseYou've already chosen the best response.0oh... I am the dumpest on the planet r^2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yes, and careful, r is the distance between centers of objects

idku
 one year ago
Best ResponseYou've already chosen the best response.0Wait so r is not that r_mars ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442545143560:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(r\) = distance between objects if the objects are spheres, then you take the distance between their centers

idku
 one year ago
Best ResponseYou've already chosen the best response.0Yes, centers (so there is an extra distance taken into account when we apply the law you mentioned, which needs to be excluded from the calculation of the attraction force).

idku
 one year ago
Best ResponseYou've already chosen the best response.0I tried to primitively apply that formula and it didn't work:( Then what task should I perform, or what formula should I use?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0what did you get for \(r\) ?

idku
 one year ago
Best ResponseYou've already chosen the best response.0I thought you completely left me. I waited bump time and asked a new question. I would have done something if I had a clue. I am so bad at physics.... i apologize.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.