Height and Acceleration of a Satellite in Orbit

[benca]

Homework Statement

A satellite is designed to orbit earth at an altitude above its surface that will place it in a gravitational field with a strength of 4.5 N/Kg

a) Calculate the distance above the surface of the earth at which the satellite must orbit

b) Assuming the orbit is circular, calculate the acceleration of the satellite and its direction

c) at what speed must the satellite travel in order to maintain this orbit?

Relevant Equations

Eg = Gme/r^2

me = mass of the earth

a)

Eg = Gme/r^2
r = √Gme/Eg
r = √[(6.67x10^-11 N*m^2*kg^2)(5.98x10^24 kg)]/(4.5 N/kg)
r = 9.41x10^6 m

h = r2 - r1
h = 9.41x10^6 m - 6.38x10^6 m
h = 3.03x10^6 m

that's over 3000 km. Did I not use for right equation? Is Eg not 4.5 N/kg?

Also for b), isn't the force of gravity the centripetal acceleration? so wouldn't it be the same equation? ac = Gme/r^2

and I don't know what it means by direction. isn't it accelerating in a circular motion? This one has really got me confused.

 

[PeroK]

that's over 3000 km. Did I not use for right equation? Is Eg not 4.5 N/kg?

Also for b), isn't the force of gravity the centripetal acceleration? so wouldn't it be the same equation? ac = Gme/r^2

and I don't know what it means by direction. isn't it accelerating in a circular motion? This one has really got me confused.

Why do you think over 3000km is wrong?

Acceletration is a vector, and has a direction. What is the directioin for uniform circular motion?

 

[benca]

Why do you think over 3000km is wrong?

Acceletration is a vector, and has a direction. What is the directioin for uniform circular motion?

Well in examples provided the satellites were much less. I guess I thought it should have been similar. And now that I say it, the direction of acceleration is towards the centre of the Earth (?)

 

[PeroK]

Well in examples provided the satellites were much less. I guess I thought it should have been similar. And now that I say it, the direction of acceleration is towards the centre of the Earth (?)

Yes. That's what "centripetal" means.

By the way, you can get the radius by noticing that the gravitational force is inversely proportional to the square of the distance. So:

$\frac{r^2}{R^2} = \frac{g_R}{g_r}$

The force you were given was just under half the Earth's surface gravity, so $r \approx 1.5 R$. $R \approx 6,000km$, so $h \approx 3,000 km$ looks about right.