Question

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?

Answer 1

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

Answer 2

@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)?

Answer 3

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)V-escape=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

Answer 4

Use the answer from a (R) to calculate the escape velocity