[Guest]

An astronaut lands on the equator of a spherical asteroid. He travels due north

100 km, without reaching the pole, then east 100 km, then south 100 km. He does

not pass the same point more than once, and finds that he is due east of his original

starting point by 200km. How many kilometres would he now need to travel by

continuing his journey in an easterly direction in order to reach his original starting

point?

(A) 200 (B) 300 © 400 (D) 500 (E) 600

Help!!!!!!!!!

[Guest]

It's surely got to be.....

600 Because the distance between two points on the same parallel is proportional to the distance of that parallel to the pole. The path along the equator is twice the length of the path along the more northern parallel, therefore, this parallel is half way to the pole. 100 is therefore 1/8 of the circumference, 200 is 1/4 of the circumference. The circumference is 800, and the distance left to cover by continuing to go east must be 600 QED

B))B))

[Guest]

Since you asked so kindly

An astronaut lands on the equator of a spherical asteroid

it means nothing, every point has infinite equators (great circle)

First of all draw the problem, it's a simple geometrical task if you have a nice figure

essentially the question is the radius of the spherical asteroid

R=300/pi [km], so the astronaut has to go 400 km into easterly direction ( C )

Edited April 11, 2010 by det

[Guest]

r*phi=100

R*phi=200

->r/R=1/2

100=pi/3*R

->R=300/pi

sorry for the disproportionate figure

Edited April 11, 2010 by det

[Guest]

"he is due east of his original starting point by 200km" so that must be the answer!!!

[Guest]

"he is due east of his original starting point by 200km" so that must be the answer!!!

But you have to continue in an easterly direction not west.

[Guest]

Er...Next time, I'll read the question properly, LOL!!!

[bonanova]

det has the right answer. Here's another way to solve it

By walking 100' to the North, the astronaut reached a latitude where a constant latitude circle is half (100/200) the length of the equator, i.e. 60^o North latitude. So 100' is 1/6 of a great circle; the equator is 600'; and having moved 200' already, exactly 400' remains.

[Guest]

det has the right answer. Here's another way to solve it

By walking 100' to the North, the astronaut reached a latitude where a constant latitude circle is half (100/200) the length of the equator, i.e. 60^o North latitude. So 100' is 1/6 of a great circle; the equator is 600'; and having moved 200' already, exactly 400' remains.

But how do you know that?

How can you tell it's 60 degrees? why not 70 or 50?

My friend and I tried to solve it for an hour and a half, tring to remember anything we have ever learned in geometry, and couldn't do it. Even after reading Bonanova's answer, we couldn't understand it.

[Guest]

But how do you know that?

How can you tell it's 60 degrees? why not 70 or 50?

My friend and I tried to solve it for an hour and a half, tring to remember anything we have ever learned in geometry, and couldn't do it. Even after reading Bonanova's answer, we couldn't understand it.

When he walked 100km north he was on a circumference with a smaller radius. he walked 100km east, and then, when he walked beack to the equator, he was 200km away from his starting position. That means that an arc of 100km in the small one is equivalent to an arc of 200km on the big one, so the small one is half the size of the big one, and so is it's radius.

Now, if you draw a right triangle from the center of the sphere(as det did) you'll find that the ratio between the small radius and the big radius is the cosine of the arc the astronaut drew by walking north. Since the ratio is 1/2, the angle is 60°, so 60° is an arc of 100km, meaning the whole equator is 100*360/60 or 600km. As he is 200km east from where he started, he can get gack there by walking 400km east or 200km west.

This is a rather confusing explanation, I admit, but I hope you understand.

[Guest]

When he walked 100km north he was on a circumference with a smaller radius. he walked 100km east, and then, when he walked beack to the equator, he was 200km away from his starting position. That means that an arc of 100km in the small one is equivalent to an arc of 200km on the big one, so the small one is half the size of the big one, and so is it's radius.

Now, if you draw a right triangle from the center of the sphere(as det did) you'll find that the ratio between the small radius and the big radius is the cosine of the arc the astronaut drew by walking north. Since the ratio is 1/2, the angle is 60°, so 60° is an arc of 100km, meaning the whole equator is 100*360/60 or 600km. As he is 200km east from where he started, he can get gack there by walking 400km east or 200km west.

This is a rather confusing explanation, I admit, but I hope you understand.

Thanks