Radius of Earth

**purplec16** · February 28th 2010, 01:41 PM

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)

**skeeter** · February 28th 2010, 01:56 PM

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)

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

**purplec16** · February 28th 2010, 02:26 PM

Why are we using sin if we have adjacent?

**skeeter** · February 28th 2010, 02:34 PM

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

**purplec16** · February 28th 2010, 02:45 PM

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

**skeeter** · February 28th 2010, 02:54 PM

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.

**purplec16** · February 28th 2010, 03:06 PM

I still dont understand what to do

**skeeter** · February 28th 2010, 04:06 PM

$h = 380$ miles

$\theta = 65.8^\circ$

solve this equation for $r$ ...

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

check if your solution is reasonable by looking up the value of the earth's radius in miles.

**Archie Meade** · February 28th 2010, 05:09 PM

hi purplec,

is your sketch like the attachment?

then, as skeeter showed

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

multiply both sides by the denominator

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

multiply out the left

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

subtract R from both sides to bring them together

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

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

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

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

factorise the R parts

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

divide both sides by the multiplier of R

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

**Drexel28** · February 28th 2010, 05:28 PM

Originally Posted by Archie Meade

hi purplec,

is your sketch like the attachment?

then, as skeeter showed

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

multiply both sides by the denominator

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

multiply out the left

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

subtract R from both sides to bring them together

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

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

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

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

factorise the R parts

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

divide both sides by the multiplier of R

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

Just a note [tex]\sin\left(65.8^{\circ}\right)[/tex] gives $\sin\left(95.8^{\circ}\right)$

**purplec16** · February 28th 2010, 05:38 PM

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

**purplec16** · February 28th 2010, 05:39 PM

I mean Archie Meade btw

**skeeter** · February 28th 2010, 06:21 PM

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

your original post ...

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?

**Archie Meade** · February 28th 2010, 06:52 PM

hi purplec,

if it's similar to this attachment,
the calculations will be the same,
as the geometry is still the same

**purplec16** · March 1st 2010, 02:52 PM

Yeah your right its like how you did it the 2nd time with out the two ends of the triangle sticking out

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