Universal gravitational constant and satellite

[hydrocarbon]

A satellite is designed to orbit Earth at an altitude above it's surface that will place it in a gravitational field with a strength of 4.5N/kg

a) calculate the distance above the surface of at which the satellite must orbit

g= GMp/r^2

r^2 = (6.67*10^-11)(5.98*10^24)/4.5 N/kg

r = sqrt(8.86*10^15) = 94147166 Meter


the answer i got was 94 million meters now that's wrong. can some please help me with the calculations? thanks

 

[D H]

Show your work, please. You made a mistake along the way but it's a bit hard to see where when you don't show the calculations.

 

[hydrocarbon]

i'm not sure how i should re-write the formula. because the answer should be a couple of

hundred km from the Earth's surface

 

[D H]

Show how you got the answer that you got: Show how you used the equation and what your intermediate steps were.

 

[hydrocarbon]

i re-wrote it again. that's for the distance

 

[D H]

You made a mistake with your exponents. What is $$10^{-11}\cdot10^{24}\;$$?

 

[hydrocarbon]

it's in the formula universal gravitational formula

 

[hydrocarbon]

A satellite is designed to orbit Earth at an altitude above it's surface that will place it in a gravitational field with a strength of 4.5N/kg

a) calculate the distance above the surface of at which the satellite must orbit

g= GMp/r^2

r^2 = (6.67*10^-11 N*m^2/kg^2)(5.98*10^24 kg)/4.5 N/kg

r = sqrt(8.86*10^15) = 94147166 Meter


the answer i got was 94 million meters now that's wrong. can some please help me with the calculations? thanks

 

[D H]

See post #6.

 

[hydrocarbon]

10^13?

 

[D H]

Bingo.

 

[hydrocarbon]

i still think the answer is wrong, over 9000 Km from the Earth's surface?

 

[D H]

You almost have the right answer. The distance that you calculated is not the distance above the surface of the Earth.

 

[hydrocarbon]

can you give me some hints on how to rearange the formula please? thanks

or am i suppose to subtract the answer from the radius of the earth? thanks again

 

[D H]

Think of it this way: You are standing on the surface of the Earth. Using the distance from the surface as r in $a=GM/r^2$ would mean you would have an enormous gravitational acceleration, and heaven forbid if you lay down inches from the surface. That r is the distance from the center of the Earth.

 

[hydrocarbon]

okay. show me were i went wrong with the calculation please

 

[D H]

I already did: "That r is the distance from the center of the Earth."

 

[hydrocarbon]

so r=9.0*10^6 - 6.38*10^6m ? because that answer got me 2.6*10^6 which is still too

much i see that you said that r is the distance from the earth. but it still doesn't tell me how

I'm calculating it incorrectly

 

[D H]

Actually, that is too little. You very first answer was off by a factor of exactly 10. You had eight significant digits in your initial answer (too many), somewhere along the line you dropped to only one significant digit (which is too few).

What makes you think you answer is wrong? Do you know the right answer?

 

[hydrocarbon]

off by a factor of 10? does that mean the answer is 900km? and if it is, how did i make such

an error (asking myself). but anyway based on the calculations the answer comes out to be

9000km unless i missed something again. do you see what i did wrong in my math?

 

[D H]

You made two errors in your original post, where you came up with 94,147,166 meters. That number was off by a factor of ten because you calculated 10^-11*10²⁴ as 10¹⁵ rather than 10¹³. Then you compounded that error by conflating the orbital radius with the orbital altitude.

Somehow you went from too many digits (94147...) to just one digit. This first appeared in post #12, where you said "over 9000 km", and later in post #18, "9.0*10^6". You are being a bit sloppy.

 

[hydrocarbon]

I wasn't getting straight answers from you. so i was just throwing up answer i did quickly hopping you would say, if it's correct or not. you tell everything else but one important thing: the answer. and you still didn't tell me if it's 9000km

 

[hydrocarbon]

and another thing. did you try the calculations yourself? and if yes. what was your answer?

 

[D H]

I wasn't answering directly because doing so is against the rules here. We are here to help you do your own homework. We do not do your homework for you.

Yes, I did the calculations myself. I always do before I help someone, and no, I'm not telling you. You need to find the answer and I will tell you something on the lines of "Yes! You got it!".

 

[hydrocarbon]

thank you very much for your service. i wasnt confident in my answers because the distance seem rediculous from answer that i saw in the examples of my course book. but once again thank you for your expertise.

 

[D H]

thank you very much for your service. i wasnt confident in my answers because the distance seem rediculous from answer that i saw in the examples of my course book. but once again thank you for your expertise.

Most of the examples in your text probably dealt with satellites in low Earth orbit (LEO). The gravitational acceleration in LEO is about 90% of that on the Earth's surface. You need to go much higher to achieve more than a 50% reduction (4.5/9.807=0.459) in the gravitational acceleration.

 

[cdaisy]

10^13?

I typed into my calculator the following : (6.67x10^-11)(5.98x10²⁴) and I got 3.98866x10¹⁴... I did it several times and came up w/ that number...

So the answer i am submitting is 1000 m than what you submitted... 24 and -11 make 13, how did i get 14?

 

[cdaisy]

thank you very much for your service. i wasnt confident in my answers because the distance seem rediculous from answer that i saw in the examples of my course book. but once again thank you for your expertise.

if you're taking it from ILC, so am I, and there are a lot of questions that don't make sense like that... the calculations and the constants are correct, but then they give you ridiculous numbers to deal with that don't make sense...