Keplers laws, weight, and revolution of planet

[jhson114]

A planet has mass Mplanet = 4.46e24 Kg and a radius Rplanet = 5900 Km
Some space explorer lands on the equator of this planet and he weighs 161lbs, whereas he weight 185lbs on Earth. He journeys over to the North pole and he finds himself weight only 95% of what he did at the equator. Now the question is how long does it take the planet to make one complete revolution. I am completely lost. how does weight of person and revolution of planet related??

 

[tony873004]

It's a bad question. There's 2 things that will cause your weight to change as a function of latitude.

1: An equatorial bulge will cause you to be further from the center of the planet when at the equator than at the pole.

2: The faster a planet spins, the lighter you will feel at the equator as the planet tries to spin you off.

In the case of Earth, the first reason is much larger.

Let's assume you're to ignore reason #1. Use the equation:

F_{c}= \frac{v^2}{r}

in this case F is your acceleration. Just figure out what v is needed to make the force decrease by (185-161lbs), and that's your rotational velocity. Then with circle formulas, and your radius, you can turn that into a period.

 

[jhson114]

i found V, and the formula i should be using is v = 2piR/T. but in equation you provided me, is F in m/s^2, V in m/s and r in meters ?

 

[tony873004]

I gave you the wrong formula, sorry.
F_{c}= \frac{v^2}{r}

should be

F_{c}= \frac{mv^2}{r}

The original way I posted it solves for acceleration, not force.

a_{c}= \frac{v^2}{r}

 

[jhson114]

weird.. I am not getting the right answer... hmm..

 

[tony873004]

... He journeys over to the North pole and he finds himself weight only 95% of what he did at the equator...

It should be the other way around. He should weigh more at the pole if the planet is spinning.

 

[tony873004]

a_1=\frac{GM}{r^2}

a_2=0.95a_1

0.95 \frac{GM }{r^2 }= \frac{v^2 } {r }

v=\sqrt{r*0.95 \frac{GM } {r^2 } }

v=\sqrt{0.95 \frac{GM } {r } }

 

[jhson114]

yeah i know the guy should weigh more at the pole but the question is messed up and says the person weighs less. anyways, i can't seem to get the right answer even with the equations from your last post.

 

[tony873004]

Plugging numbers into the final equation I gave you, I get 6921 m/s which seems reasonable if they goofed up and should have said he weighs less at the equator. Remember to convert km to meters for radius.

Does the back of the book give an answer?

 

[jhson114]

thats what i got too. so i took velocity and plugged it into v = 2piR/T and solved for T which should give me the period. it gives it to me in seconds, and i need the answer in hours, so i covert it but i get a really small number between 1-2.

 

[jhson114]

its one of those internet questions where you input answer and it simply tell you whether you got it right or wrong.

 

[tony873004]

where did you get velocity?

Copy and paste the question directly from the online homework problem

 

[jhson114]

A space explorer comes across a planet with a small moon in orbit around it. He observes that the moon completes one full orbit every 30 days and moves at a distance R = 370 million m from the center of the planet. What is the mass of the planet?

Mplanet = 4.46e24 kg


Upon further exploration he finds another satellite orbiting the planet in one third the time the moon did. What is the orbital radius of the satellite?

Rsatellite = 178 million m


The explorer now lands his spacecraft on the surface of the planet near the equator. He weighs himself and finds he weighs only Wplanet = 161 lbs while on Earth he weighed Wearth = 185 lbs.

Assuming the explorer hasn't dieted on his trip, what is the radius of the planet?

Rplanet = 5900 km


The explorer continues his studies and soon ends up at the North pole. There he finds he weighs only 95% of what he did at the equator. How long does it take the planet to make one complete revolution?

Trotation = ___ hours

 

[tony873004]

I get 1.49 hours. Hopefully this isn't due tommorow! This isn't that hard of a question. It probably has something to do with their typo that you'd weigh more at the equator.

Is this Webassign.com? I used them last semester, and e-mailed them when I found a typo that yielded the wrong answer. They corrected it and thanked me.

 

[jhson114]

1.49 is incorrect. this is the number i got too. its due wednesday. not webassing, but similar. its called tycho.

 

[jhson114]

can anyone else help me with this problem? this one is driving me nuts