Help with a Space question? (1 Viewer)

[Aluminesis]

Hey guys, I'm doing past half-yearly papers and I got stumped on a question. Hoping I can get some help here.

Question:
[FONT=&quot]In the launch of a particular satellite, the satellite was release from a rocket such that it moved into a stable orbit around the Earth. After the satellite had completed a number of orbits of Earth, each taking 90 minutes, onboard rockets were used to propel the satellite into a much more distant stable orbit, with a radius 10 times larger than the original. Based on this information, which of the following could be closest to the orbital period of the satellite in the final more distant orbit?[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot](A) 47.4 hours[/FONT]
[FONT=&quot](B) 38.7 hours [/FONT]
[FONT=&quot](C) 22.5 hours [/FONT]
[FONT=&quot](D) 15 hours [/FONT]

The answer is A. I tried using Kepler's Law of Periods but it didn't work out. Am I approaching it from the wrong angle or am I on the right track but making silly mistakes?

I'd really appreciate it if someone could post the full working out involved with this question. Thanks!

 

[laplace4]

That is so easy, you might as well just give up on hsc altogether if you cant do that

 

[Aluminesis]

I'm sure it is, mate, but not every person who can't tie their shoelace goes and chucks their shoe out the window so I'll take my chances

 

[laplace4]

nah you are a no hoper lol

use the keplar eqn to find the initial distance from the centre of the earth, then multiply that value by 10, then sub that back ito the eqn and solve the period

there are only two unknowns in the equation , r and T , if you know one you can find the other

 

[OmmU]

Answer is A.

I suspect you are having some calculation errors:

My working:

Find out the initial orbital radius:

Plug your info into the formula:


Rearrange and don't forget to cube root to get r. (Since only goes to like 4dp, wrote down to 4dp.

Then plug 10r into the equation and rearrange to get :




= 47.4hrs

 

[laplace4]

lol you are a noob too, you didnt even need constants to do it

 

[laplace10]

lol you dont even need the constants or anything

you can just simplify everything down to final period = sqrt ( 10^3 x T^2 ) , where T is the first period.

Thats just by pure algberaic manipulation

 

[Fizzy_Cyst]

Hey guys, I'm doing past half-yearly papers and I got stumped on a question. Hoping I can get some help here.

Question:
[FONT=&quot]In the launch of a particular satellite, the satellite was release from a rocket such that it moved into a stable orbit around the Earth. After the satellite had completed a number of orbits of Earth, each taking 90 minutes, onboard rockets were used to propel the satellite into a much more distant stable orbit, with a radius 10 times larger than the original. Based on this information, which of the following could be closest to the orbital period of the satellite in the final more distant orbit?[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot](A) 47.4 hours[/FONT]
[FONT=&quot](B) 38.7 hours [/FONT]
[FONT=&quot](C) 22.5 hours [/FONT]
[FONT=&quot](D) 15 hours [/FONT]

The answer is A. I tried using Kepler's Law of Periods but it didn't work out. Am I approaching it from the wrong angle or am I on the right track but making silly mistakes?

I'd really appreciate it if someone could post the full working out involved with this question. Thanks!

Seeing as it is still orbiting around the same object r^3/T^2 is constant.

T^2 final = (r^3final * T^2 initial)/r^3 initial


T^2 final = (r^3final/r^3 initial) * T^2 initial

T^2 final = (r final / r initial)^3 * T^2 initial

T^2 final = (10)^3 * 90^2

T final = sqrt (1000*8100)

T final = 2846mins

T final = 47.43hrs

 

[jyu]

Seeing as it is still orbiting around the same object r^3/T^2 is constant.

T^2 final = (r^3final * T^2 initial)/r^3 initial


T^2 final = (r^3final/r^3 initial) * T^2 initial

T^2 final = (r final / r initial)^3 * T^2 initial

T^2 final = (10)^3 * 90^2

T final = sqrt (1000*8100)

T final = 2846mins

T final = 47.43hrs

wording of the question
10 times longer than the original means 11 times the original?

 

[Fizzy_Cyst]

True, that could definitely be debated!

But then the answer would be greater than 47.43, prolly closer to 55. Seeing as that is not an option in the answers, the answer must be 47.43 =)