# Speed of satellite

1. Mar 18, 2005

### dalitwil

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!

2. Mar 18, 2005

### BobG

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.

$$\sqrt{\frac{GM}{r}}$$

3. Mar 18, 2005

### Staff: Mentor

This equation is incorrect. Derive the correct equation using Newton's law of gravity and Newton's 2nd law of motion.

4. Mar 18, 2005

### xanthym

{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. Mar 20, 2005

### tony873004

$$\sqrt{\frac{2GM}{r}}$$