Scatch space mod prob. is need a formula and answer check

  • Thread starter Thread starter cowgiljl
  • Start date Start date
  • Tags Tags
    Formula Space
Click For Summary
The discussion focuses on calculating the speed of a command module orbiting the Moon and the time it takes for one revolution. The initial speed calculation yielded 1634 m/s, which was confirmed as correct. However, there was confusion regarding the period calculation, with one participant using Kepler's 3rd law and arriving at a period of 118 minutes, while another suggested a simpler method by dividing the orbit's circumference by the speed. Ultimately, the importance of verifying calculations and considering alternative approaches to problem-solving was emphasized. The conversation highlights the collaborative nature of problem-solving in physics.
cowgiljl
Messages
63
Reaction score
1
i think i got it right it wants to know the speed of the command module and the time it takes to do 1 revolution
mass moon = 7.35E22 kg
radius moon = 1.74E6 m
altitude 60 mi

fromula used was Vcm^2 = GMm/Ro

the speed of the module was 1634 m/s

The time formula thans to enigma was

P = 2\pi \sqrt{\frac{a^3}{\mu}}}
T = 6511 seconds/60
T = 108 min

is this right?
 
Physics news on Phys.org
Your speed looks correct, using your value for mass of the moon.

But I don't follow what you're doing to get the period.

I'm using Kepler's 3rd law:
T^2 = (\frac{4\pi^2}{GM_m})r^3
and here r = (1.74 x 106 + 1609 x 60) so using G=6.67x10-11 and Mm=7.35x1022 I'm getting
T = 7063 s or 118 min.

(Incidentally, my book lists the mass of the moon as 7.36x1022 kg, but what's a mere 100,000,000,000,000,000,000 kg among friends?)
 
you re right gnome

I wrote down and typed the wrong answer but my calculations were correct. the mind was quicker than the pen

thanks again
 
Sometimes its good to step back & think about what we're doing.

While I was walking my dog, I was wondering why we're using that crazy formula to get the time of 1 revolution. (Heresy -- calling Kepler's 3rd law a crazy formula! )

We already know the speed. Just divide the circumference of the orbit by the speed & get the same answer.]
 
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 ·
Replies
2
Views
2K
Replies
3
Views
6K
  • · Replies 2 ·
Replies
2
Views
10K