Asked by Ana
A synchronous satellite, which always remains above the same point on a planet's equator, is put in orbit around Neptune so that scientists can study a surface feature. Neptune rotates once every 16.1 h. Use the data of Table 13.2 to find the altitude of the satellite.
Mass of Neptune=1.03*10^26 kg
Radius of Neptune=2.21*10^7m
{(6.67*10^-11)(1.03*10^26)(57960)^2/{4pi^2)}^(1/3)=83615480.89
and then I subtracted that answer from 2.21*10^7=61515480.89
when i insert 61515480.89 as my answer homework site says "Your response is off by a multiple of ten". What am I missing?
Thanks
Mass of Neptune=1.03*10^26 kg
Radius of Neptune=2.21*10^7m
{(6.67*10^-11)(1.03*10^26)(57960)^2/{4pi^2)}^(1/3)=83615480.89
and then I subtracted that answer from 2.21*10^7=61515480.89
when i insert 61515480.89 as my answer homework site says "Your response is off by a multiple of ten". What am I missing?
Thanks
Answers
Answered by
Damon
I agree with orbit radius = 8.361*10^7
I think the radius of Neptune is more like 2.48 * 10^7 meters.
so I get 5.88 * 10^7
I think the radius of Neptune is more like 2.48 * 10^7 meters.
so I get 5.88 * 10^7
Answered by
bobpursley
I agree, you are off by a factor of 10.
8.3E7-2.2E7=6.2E7
8.3E7-2.2E7=6.2E7
Answered by
Damon
I bet they want the answer to be in the form
5.88 * 10^7 meters,
to the same significant figures as given in your table (of course I also do not agree with their planet radius)
5.88 * 10^7 meters,
to the same significant figures as given in your table (of course I also do not agree with their planet radius)
Answered by
Damon
I do not understand this off by a factor of ten. Ana's answer, 61515480.89 = 6.15*10^7 approx., looks fine to me except that I disagree about the planet radius.
Answered by
Ana
I inserted 6.2E7 and i also try to do 62000000 but it says the same thing. How else can I insert my answer?
Answered by
Damon
Try 6.15E7
and try
5.88E7
and try
5.88E7
Answered by
Damon
Three of us have done this independently now (I started with my own planet data) and bob and I agree with you.
Answered by
Ana
I appreciate who ever is helping me out, but again I tried 6.15E7 and 5.88E7 still it isn't working out. I can only insert answer two times after that it wont work.
I don't know what else I can do
I don't know what else I can do
Answered by
Damon
Nor do I. However the august text had the radius of the planet wrong (or at least not the same as my data) so it is possible that they have the answer wrong. That is all I can think of.
Answered by
Ana
the mass and radius were given by professor so I just followed that.
maybe you are right.
thanks anyway
maybe you are right.
thanks anyway
Answered by
Damon
ps,
It is two physicists helping you out.
It is two physicists helping you out.
Answered by
Ana
I believe you.
Answered by
Core'
i know this is WAY over due but your equation is off and a figure of 10-3 is unaccounted for. your answer should be 6.15e4
Answered by
CandyLand
Core', even though your answer is WAY overdue, you still helped me out tremendously. I have been looking for this answer for hours now and you just relieved a lot of stress from me, so thank you for posting this answer!!!
Answered by
Tutor Allan
I have solution in pdf but cant upload here. eyehelp @ gmail com (no spaces and remember to put a dot before com)
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