eyeseeyou
Well-Known Member
idkIs my answer right?
idkIs my answer right?
.How do I calculate the period
Also does Period represent gravitational constant in the orbital velocity formula?
lol waste of time doing it when there's no solutions....
I did a similar question today but a bit harder in its wording..I applied the same technique..so I think it is correct.lol waste of time doing it when there's no solutions....
How's it a waste of time?lol waste of time doing it when there's no solutions....
Because we could've done it completely wrong and not know lolHow's it a waste of time?
Why don't you just think about the questions before posting? You look at what you have and what you want and see which equation(s) to use. There's a reason why people who don't do maths/general can still do HSC physics - there are no tricks involved in calculations. Maybe self learning isn't working out so well?How's it a waste of time?
What?r=2.5x10^6m + the altitude given
If we rearrange the ΔGPE formula we get E(p)=Gm1m2(1/r(i) - 1/r(f))
∴1x10^7=Gm1m2(1/(2.6x10^6) - 1/(2.7x10^6))
∴Gm1m2=7.02x10^14
Now we calculate the ΔGPE for the fall by subbing in Gm1m2 as 7.02x10^14
∴ΔGPE=7.02x10^14(1/(2.7x10^6) - 1/(2.66x10^6))
∴ΔGPE=-4.9x10^6 J
10 MJ = 10 x 10^6 JoulesWhat?
Thanks10 MJ = 10 x 10^6 Joules
= 1 x 10^7 Joules
Why'd u switch the numbers around?r=2.5x10^6m + the altitude given
If we rearrange the ΔGPE formula we get E(p)=Gm1m2(1/r(i) - 1/r(f))
∴1x10^7=Gm1m2(1/(2.6x10^6) - 1/(2.7x10^6))
∴Gm1m2=7.02x10^14
Now we calculate the ΔGPE for the fall by subbing in Gm1m2 as 7.02x10^14
∴ΔGPE=7.02x10^14(1/(2.7x10^6) - 1/(2.66x10^6))
∴ΔGPE=-4.9x10^6 J
To account for the negative in -(Gm1m2) so to make it positive we do r(i)-r(f) instead of r(f)-r(i)Why'd u switch the numbers around?
Oh right rightTo account for the negative -(Gm1m2) so to make it positive we do r(i)-r(f) instead of r(f)-r(i)