1. ## Radius of Earth

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)

2. 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)
$\sin{\theta} = \frac{r}{r+h}$

3. Why are we using sin if we have adjacent?

4. 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

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

6. 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.

7. I still dont understand what to do

8. $h = 380$ 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.

9. 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

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

10. 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 $$\sin\left(65.8^{\circ}\right)$$ gives $\sin\left(95.8^{\circ}\right)$

11. 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

12. I mean Archie Meade btw

13. 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?

14. hi purplec,

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

15. 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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