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• Feb 28th 2010, 01:41 PM
purplec16
A spacelab circles Earth at an altitude of 380 miles. When an astronaut views the horizon of Earth the angle theta is 65.8 degrees. Use this information to estimate the radius of Earth.

What do I do?
(I know how it looks so you dont have to draw it but I dont know what am finding, hence I cant find it)
• Feb 28th 2010, 01:56 PM
skeeter
Quote:

Originally Posted by purplec16
A spacelab circles Earth at an altitude of 380 miles. When an astronaut views the horizon of Earth the angle theta is 65.8 degrees. Use this information to estimate the radius of Earth.

What do I do?
(I know how it looks so you dont have to draw it but I dont know what am finding, hence I cant find it)

$\displaystyle \sin{\theta} = \frac{r}{r+h}$
• Feb 28th 2010, 02:26 PM
purplec16
Why are we using sin if we have adjacent?
• Feb 28th 2010, 02:34 PM
skeeter
Quote:

Originally Posted by purplec16
Why are we using sin if we have adjacent?

better take another look at your sketch.

the adjacent side is the distance from the astronaut to the point where the horizon touches the earth's surface.

the hypotenuse is (r+h)

the opposite side is r
• Feb 28th 2010, 02:45 PM
purplec16
So what am I solving for h or r?, r right? so dont i have to find h? whats h?
• Feb 28th 2010, 02:54 PM
skeeter
Quote:

Originally Posted by purplec16
So what am I solving for h or r?, r right? so dont i have to find h? whats h?

the height of the spacecraft above the earth's surface.
• Feb 28th 2010, 03:06 PM
purplec16
(Worried) I still dont understand what to do
• Feb 28th 2010, 04:06 PM
skeeter
$\displaystyle h = 380$ miles

$\displaystyle \theta = 65.8^\circ$

solve this equation for $\displaystyle r$ ...

$\displaystyle \sin{\theta} = \frac{r}{r+h}$

check if your solution is reasonable by looking up the value of the earth's radius in miles.
• Feb 28th 2010, 05:09 PM
hi purplec,

is your sketch like the attachment?

then, as skeeter showed

$\displaystyle Sin(65.8^o)=\frac{opposite\ to\ angle}{hypotenuse}=\frac{R}{R+380}$

multiply both sides by the denominator

$\displaystyle (R+380)Sin(65.8^o)=\frac{R(R+380)}{R+380}=R$

multiply out the left

$\displaystyle RSin(65.8^o)+380Sin(65.8^o)=R$

subtract R from both sides to bring them together

$\displaystyle RSin(65.8^o)-R+380Sin(65.8^o)=R-R=0$

subtract the 380Sin part from both sides to have the R terms alone on the left

$\displaystyle RSin(65.8^o)-R+380Sin(65.8^o)-380Sin(65.8^o)=-380Sin(65.8^o)$

$\displaystyle RSin(65.8^o)-R=-380Sin(65.8^o)$

factorise the R parts

$\displaystyle R[Sin(65.8^o)-1]=-380Sin(65.8^o)$

divide both sides by the multiplier of R

$\displaystyle R=\frac{-380Sin(65.8^0)}{Sin(65.8^o)-1}=380\frac{(-1)Sin(65.8^o)}{(-1)[1-Sin(65.8^o)]}=380\frac{Sin(65.8^o)}{1-Sin(65.8^o)}$
• Feb 28th 2010, 05:28 PM
Drexel28
Quote:

hi purplec,

is your sketch like the attachment?

then, as skeeter showed

$\displaystyle Sin(65.8^o)=\frac{opposite\ to\ angle}{hypotenuse}=\frac{R}{R+380}$

multiply both sides by the denominator

$\displaystyle (R+380)Sin(65.8^o)=\frac{R(R+380)}{R+380}=R$

multiply out the left

$\displaystyle RSin(65.8^o)+380Sin(65.8^o)=R$

subtract R from both sides to bring them together

$\displaystyle RSin(65.8^o)-R+380Sin(65.8^o)=R-R=0$

subtract the 380Sin part from both sides to have the R terms alone on the left

$\displaystyle RSin(65.8^o)-R+380Sin(65.8^o)-380Sin(65.8^o)=-380Sin(65.8^o)$

$\displaystyle RSin(65.8^o)-R=-380Sin(65.8^o)$

factorise the R parts

$\displaystyle R[Sin(65.8^o)-1]=-380Sin(65.8^o)$

divide both sides by the multiplier of R

$\displaystyle R=\frac{-380Sin(65.8^0)}{Sin(65.8^o)-1}=380\frac{(-1)Sin(65.8^o)}{(-1)[1-Sin(65.8^o)]}=380\frac{Sin(65.8^o)}{1-Sin(65.8^o)}$

Just a note $$\sin\left(65.8^{\circ}\right)$$ gives $\displaystyle \sin\left(95.8^{\circ}\right)$
• Feb 28th 2010, 05:38 PM
purplec16
no its not I tried to upload it but it wont
its the earth then its this triangle forming at the edge of it, and it bottom length is 360 mi and it tips the earth then the line extends and to the center of the earth and thats what their tellin me to find and the angle touching it is 68.5 degrees
• Feb 28th 2010, 05:39 PM
purplec16
• Feb 28th 2010, 06:21 PM
skeeter
Quote:

Originally Posted by purplec16
no its not I tried to upload it but it wont
its the earth then its this triangle forming at the edge of it, and it bottom length is 360 mi and it tips the earth then the line extends and to the center of the earth and thats what their tellin me to find and the angle touching it is 68.5 degrees

Quote:

A spacelab circles Earth at an altitude of 380 miles. When an astronaut views the horizon of Earth the angle theta is 65.8 degrees. Use this information to estimate the radius of Earth.
so ... which is correct?
• Feb 28th 2010, 06:52 PM