Speed of satellite

  • Thread starter dalitwil
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  • #1
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A typical GPS (Global Positioning System) satellite orbits at an altitude of 2.0e7 m.

Find the orbital speed of such a satellite in km/s.


So we know:

Mass of the earth (M) =5.97e24
Radius of the earth (R) =6.37e6
G is constant =6.67e-11

I used the equation:

Final velocity=square root (2GM/R)

square root ((2*6.67e-11*5.97e24)/(6.37e6+2.0e7))

And dividing the answer by 1000 to obtain km/s I get 5.50km/s.

The answer is 3.9km/s.... I think my error may be related to adding the altitude to the radius of the earth at the bottom of the equation, but I'm not sure.

Thanks guys!
 

Answers and Replies

  • #2
BobG
Science Advisor
Homework Helper
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Yes, that's part of your problem. What's the extra 2 for in the numerator?

GM=3.986004418e14 (3.986e14 gets you more than close enough for this problem) meters (your way works too, but why multiply two constants together over and over). BTW, if you want smaller numbers, convert to km right off the bat, making GM=3.986e5.

You have to add in the radius of the Earth to convert altitude to radius.

Your equation for speed is:

[tex]\sqrt{\frac{GM}{r}}[/tex]
 
  • #3
Doc Al
Mentor
44,994
1,268
dalitwil said:
I used the equation:

Final velocity=square root (2GM/R)
This equation is incorrect. Derive the correct equation using Newton's law of gravity and Newton's 2nd law of motion.
 
  • #4
xanthym
Science Advisor
410
0
dalitwil said:
A typical GPS (Global Positioning System) satellite orbits at an altitude of 2.0e7 m.

Find the orbital speed of such a satellite in km/s.


So we know:

Mass of the earth (M) =5.97e24
Radius of the earth (R) =6.37e6
G is constant =6.67e-11

I used the equation:

Final velocity=square root (2GM/R)

square root ((2*6.67e-11*5.97e24)/(6.37e6+2.0e7))

And dividing the answer by 1000 to obtain km/s I get 5.50km/s.

The answer is 3.9km/s.... I think my error may be related to adding the altitude to the radius of the earth at the bottom of the equation, but I'm not sure.

Thanks guys!
{Radius of Earth} = R = {6.37e(+6) m}
{Mass Of Earth} = M = {5.97e(+24) kg}
{Gravitational Constant} = {6.67e(-11) N*m2*kg2}
{Mass of Satellite} = m
{Gravitational Force at Altitude (2.0e(+7) m)} = G*M*m/{R + 2.0e(+7)}2 =
= {6.67e(-11)}*{5.97e(+24)}*m/{6.37e(+6) + 2.0e(+7)}2 =
= (0.5726)*m

{Centripetal Force} = m*v2/{6.37e(+6) + 2.0e(+7)} =
= m*v2/{26.37e(+6)}

At equilibrium:
{Centripetal Force} = {Gravitational Force at Altitude (2.0e(+7) m)}
::: ⇒ m*v2/{26.37e(+6)} = (0.5726)*m
::: ⇒ v2 = (0.5726)*{26.37e(+6)}
::: ⇒ v = (3886 m/sec) = (3.886 km/sec)


~~
 
  • #5
tony873004
Science Advisor
Gold Member
1,751
143
Your original equation
[tex]\sqrt{\frac{2GM}{r}}[/tex]
is the equation for escape velocity. You probably just mixed the two up.
 

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