physics
posted by Jets on .
A satellite with mass 6.00*10^3 is in the equatorial plane in a circular orbit. The planet's mass = 6.59 *10^25 and a day of length is 1.6 earth days. How far from the center (in m) of the planet is the satellite? What is the escape velocity (km?sec) from the orbit?

a. 1.29*10^8; b. 8.2547 km/sec

@frank could you tell the steps????
A satellite with a mass of ms = 4.00 × 103 kg is in a planet's equatorial plane in a circular "synchronous" orbit. This means that an observer at the equator will see the satellite being stationary overhead (see figure below). The planet has mass mp = 6.13 × 1025 kg and a day of length T = 1.3 earth days (1 earth day = 24 hours).
How far from the center (in m) of the planet is the satellite?
What is the escape velocity (in km/sec) of any object that is at the same distance from the center of the planet that you calculated in (a)? 
Geosynchronous Orbit Equation:
R^3=(T^2)*(G)*(M_p)/(4)*(pi^2)
Time is in seconds
T=1.6 earth days=38.4hrs=13824seconds or
1.38*10^4 seconds
this the equation i I used to get my answer.
b)Vescape=sqrt(2)*(m_planet)*G/R
to do b you need a to solve; do not forget to take cubed root for a. Everybody has different variables.
I have trouble with Bungee jumper 
Use the answer from a (R) to calculate the escape velocity