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)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
Use the answer from a (R) to calculate the escape velocity
To find the distance of the satellite from the center of the planet, we can use the formula for the gravitational force between two objects:
F = G * (m1 * m2) / r^2
Where F is the gravitational force, G is the gravitational constant (approximately 6.674 × 10^-11 N m^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects.
In this case, the satellite and the planet are in a circular orbit, so the gravitational force is equal to the centripetal force:
F = (m_satellite * v^2) / r
Where m_satellite is the mass of the satellite, v is its velocity in the orbit, and r is the distance from the center of the planet.
Setting these two equations equal to each other, we get:
G * (m1 * m2) / r^2 = (m_satellite * v^2) / r
Rearranging the equation and solving for r, we get:
r = (G * m1 * m2) / (m_satellite * v^2)
Now let's calculate the value. Given:
Satellite mass (m_satellite) = 6.00 * 10^3 kg
Planet mass (m1) = 6.59 * 10^25 kg
Orbital velocity (v) = 2πr / T, where r is the distance from the center of the planet and T is the period of the orbit. In this case, T is 1.6 Earth days, or 1.6 * 24 * 60 * 60 seconds.
Plugging in the values, we can calculate r:
r = (6.674 × 10^-11 N m^2/kg^2 * (6.59 * 10^25 kg) * (6.00 * 10^3 kg)) / ((2πr / (1.6 * 24 * 60 * 60))^2)
To solve this equation, we can use an iterative method, such as the Newton-Raphson method, to find the value of r that satisfies the equation. However, it is worth noting that since r is in both the numerator and denominator of the equation, we cannot solve for r analytically. Therefore, numerical methods are required to obtain an approximate solution.
Now, let's move on to calculating the escape velocity. The escape velocity is the velocity required for an object to escape the planet's gravitational field, i.e., the minimum velocity required to move infinitely far away from the planet.
The escape velocity can be calculated using the formula:
v_escape = √(2 * G * m1 / r)
Plugging in the values:
v_escape = √(2 * 6.674 × 10^-11 N m^2/kg^2 * (6.59 * 10^25 kg) / r)
We can use the value of r obtained from our previous calculation to find the escape velocity.