# Need Check On Orbital Satellite Problem

#### tachu101

1. Homework Statement
A 347 kg satellite orbits mercury and makes 6.6 orbits in 1 day.

Mercury ---- Radius 2.44*10^6 meters ----- Mass 3.27*10^23 kg

1.Find the Period of Revolution
2.Calculate the Tangential Velocity
3.Calculate the height above the surface of the planet that the satellite is in orbit
4.Calculate the force of gravity at a distance of 1.50*10^6 meters

2. Homework Equations
Planetary motion Equations

3. The Attempt at a Solution

1. Period= 24 hours/6.6 orbits ---- 13090.90909 seconds (per revolution)

2. Tangential Velocity= (2*pi*radius)/Period ----- (2*pi*(2.44*10^6))/13090.90 second ---- Vt= 1171 m/s

3. My friend said that this was the equation for the third part, does anyone know if this is right.
h=((Gravity Constant*Mass Mercury)/Vt^2) - radius of Mercury -----
h=((6.67*10^-11)(3.27*10^23)/1171^2) - 2.44*10^6 meters -----
h= 13465956 meters or 1.3465*10^7 meters (is this right)!?!

4. Same friend said that the force of gravity would be :
g= (6.67*10^-11)(3.27*10^23)/ ((2.44*10^6+1.5*10^6)^2)
g= 1.405 m/sec^2

Can some one check these answers because I just feel like I did something wrong.

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#### dotman

Hello,

You've used the radius to represent two different quantities.

When you calculated the velocity in part (2), you used the radius 2.44*10^6 m as the radius of the orbit. However, when you calculated the height in part (3), you used the radius of 2.44*10^6 m as the radius of mercury. I'm not sure which it's supposed to be, but it can't be both.

#### tachu101

Can anyone else help?

#### Dick

Homework Helper
Listen to dotman first. Your result for the tangential velocity is correct only if the radius of Mercury is the same as the radius of the orbit. You need to use 'Gravity constant' from the beginning. GM/r^2=v^2/r. v=2*pi*r/T. T is period, M is mass of Mercury and r is the radius of the orbit. The radius of Mercury doesn't matter much.

#### tachu101

I redid the questions

1. Period= 24 hours/6.6 orbits ---- 13090.90909 seconds (per revolution)

2. Book says : ((G*MassMercury*Period^2)/4*pi^2)^(1/3) =Radius of Orbit -- 4.55*10^6 meters?
Then I put into ---- Tangential Velocity= (2*pi*radius)/Period --- 2.1875*10^3 m/s (too fast)?

3. Then height above the surface of the earth is
Height = Radius Of orbit - Radius of Mercury ---- Height = 2.117*10^3 meters

4. The gravity at 1.50*10^6 meters away I think I have right.

Can someone tell me if I am doing this right?

#### Dick

Homework Helper
You mean 10^6 in the height, right? Not 10^3. Otherwise, looks great!

#### tachu101

Yea, typing error

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