## idku one year ago Advanced PHYSICS question. SATELLINE in SPACE (((FORCES, ACCELERATIONS etc...)))

1. idku

Scientists 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| $$\large \rm m_{mars} = 6.4191 \times 10^{23} kg$$ $$\large \rm r_{mars} = 3.397 \times 10^6 m$$ $$\large \rm G = 6.67428 \times 10^{-11} {~}(N-m^2)/kg^2$$

2. idku

1) What is the force of attraction between Mars and the satellite?

3. idku

((I dissapear from the site, due to this site's faultiness when it comes to connecting. Apologize ))

4. ganeshie8

you may use newton's law of gravity : $F = G \dfrac{m_1m_2}{r^2}$

5. idku

G × (2700kg × 6.4191×10$$^{23}$$kg) / (3.397 × 10$$^{6}$$m) and I will plug my G of course...

6. idku

3.41 N is what I get.

7. idku

wrong_:

8. idku

6.67428×10^(-11)• (2700 × 6.4191×10^(23)) / (3.397 × 10^(6)) this is what I plugged into wolfram, and got 3.41

9. idku

oh... I am the dumpest on the planet r^2

10. ganeshie8

Yes, and careful, r is the distance between centers of objects

11. idku

Wait so r is not that r_mars ?

12. ganeshie8

|dw:1442545143560:dw|

13. ganeshie8

$$r$$ = distance between objects if the objects are spheres, then you take the distance between their centers

14. idku

Yes, 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).

15. idku

I tried to primitively apply that formula and it didn't work-:( Then what task should I perform, or what formula should I use?

16. ganeshie8

what did you get for $$r$$ ?

17. idku

I 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.