Recent content by jantdroid

∠⊆⊆⊕⊕→ℝℝβ Nice! Thanks for giving me this technique on making equations neat, I'll study how to get that quartic equation right away. I'll get back to you once I've confirmed the answer. Thank you! :D

Here, me trying to apply your method, Let F> 0.001 0.001 < G (D-x)2 - 2G (x2) / (D-x)2 (x2) 0.001 < G (D2- 2Dx +x2) - 2Gx / (D2-2Dx+x2) (x2) 0.001 < GD2 - 2GDx + Gx2 -2Gx2/ X2D2 - 2Dx3 + x4 0 < GD2- 2GDx - Gx2 / x2D2- 2Dx3 + x4 - 0.001 Finding the LCD: 0 < ( GD2- 2GDx -Gx2 - 0.001 D2x2...

I don't quite get where the "H" came from and what exactly is the answer (distance) from planet a to b

F=FA−FB Since the force needed is greater than 0.001N 0.001< 3000/da2 - 6000/db2 transposing the 0.001 value to the right side of the equation, 0< 3000/da2 - 6000/db2 - 0.001 Finding the LCD of the solution 0< ( 3000db2-6000da2-0.001da2db2 ) db2 da2 Given that db= 6 000 000 - da then...

Yup, my classmate tried to solved it too still get the 4th degree equation. lol

Yah, the equation is on the 4th degree. Not sure how to proceed with it by picking out the critical points and do some test points. :3

The one that says find the distance(s) where gravitational force is above 0.001 N

I think so, the question kinda confuses me. "What are the distances (ranges) from Planet A to B where the gravitational force is above 6 000 000 and I'm concerned about the rocket in the middle too.

Homework Statement Given that the force of gravitation between Planet A (the one in the left side of the drawing), Fa=3000/da2 and the force of gravitation between Planet B and the rocket, Fb= 6000/ db2. Assuming that the three bodies involved is in stationary. What are the distances (ranges)...