Recent content by theblazierbroom

I think this is it! Thanks for the help.

Ah, I see what happened. The kinetic energy to escape the orbit of the sun has to equal the potential energy of the circular orbit, not just the additional energy. $$ T = U_{circular \, orbit} $$ $$ T = m v_{orbit}^{2} $$ Adjusting to have the frame of reference of the sun, and knowing that...

I see the contradiction and what's been throwing me off. Then, I have that: ##mgR_{e}## ... The minimum energy required to escape Earth's gravitational pull ##\frac{1}{2}mv_{orbit}^{2}## ... The minimum energy required to escape Sun's gravitational pull ##\frac{1}{2}mv_{f}^2## ... The final...

Thanks for the outline! The kinetic energy needed to escape Earth's gravity is: $$ \frac{1}{2} m v_{escape}^{2} = m g R_{e} $$ It will be in a circular orbit around the Sun, with relative velocity around sun = ##\textbf{v}_{escape} + 18.5 \, mi \, s^{-1}##. Using the fact that Potential...

Okay, so now I'm thinking, regardless of what launch direction it is, the launch probe must have with a minimum of 12.1 ##mi \, s^{-1}## to leave Earth's gravitational pull. The issue is breaking out of the orbit around the sun. If launching in the direction of orbit, launch with 12.1+18.5 =...

For the (a) portion of the problem, it asks to calculate the minimum speed a probe must be launched from earth to escape the solar system with residual speed of 10 ##mi \, s^{-1}## relative to the sun. To find the minimum speed, I assumed the gravitational force affecting the probe by the sun...

I see, you helped me get out. I think some good amount of overthinking was occurring in my head haha Thanks!

Wait, so set ## \Delta v ## / ##\Delta t## = - (k/m) * v? So then if I do ## \Delta v ## = - (k/m) * v * ## \Delta t ##, then I set v in the RHS as the velocity from the previous interval?

When I used differential equation techniques, I found the function of x and v to be a negative exponential function. However, based on the notes, I believe the problem wants me to use finite summations as the relevant equations above. This stumps me because the acceleration is dependent on the...