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a. Draw and label the velocity and acceleration vectors of the shuttle at point P.
The velocity vector is tangent to the orbit at point P, and the acceleration vector points directly towards the center of the Earth.
b. Write the equation that can describe the gravitational force on the shuttle.
The gravitational force on the shuttle is given by the following equation:
[tex]F_g = G * (m_earth * m_shuttle) / (r_earth + h)^2[/tex]
where:
* G is the universal gravitational constant
* m_earth is the mass of the Earth
* m_shuttle is the mass of the shuttle
* r_earth is the radius of the Earth
* h is the height of the shuttle above the surface of the Earth
c. Derive the equation for the shuttle's acceleration when it is in orbit.
The shuttle's acceleration in orbit is given by the following equation:
[tex]a = F_g / m_shuttle[/tex]
Substituting the equation for the gravitational force, we get:
[tex]a = G * (m_earth * m_shuttle) / (r_earth + h)^2 * 1 / m_shuttle[/tex]
Simplifying, we get:
[tex]a = G * m_earth / (r_earth + h)^2[/tex]
This equation shows that the shuttle's acceleration is independent of its mass. All satellites in orbit around the Earth experience the same acceleration, regardless of their mass.
d. Derive the equation for the velocity of the satellite as it stays in orbit.
The velocity of the satellite as it stays in orbit is given by the following equation:
[tex]v = sqrt(G * m_earth / (r_earth + h))[/tex]
This equation shows that the satellite's velocity is proportional to the square root of the Earth's mass and inversely proportional to the square root of the distance between the satellite and the center of the Earth.
e. How will the velocity change if the shuttle increases its height above the surface? Explain.
If the shuttle increases its height above the surface, its velocity will decrease. This is because the gravitational force on the shuttle will decrease as the distance between the shuttle and the center of the Earth increases.
f. If the shuttle's period is synchronized with that of Earth's rotation, what is the height of the shuttle? [tex](1 day = 8.64x10^4s, ME = 6x10^24kg, RE = 6.4x10^6m)[/tex]
The period of a satellite in orbit is the time it takes to complete one orbit. The period of a satellite in geostationary orbit, which is an orbit that is synchronized with the Earth's rotation, is 24 hours.
The height of a satellite in geostationary orbit can be calculated using the following equation:
[tex]h = sqrt[G * ME / (RE * omega^2)] - RE[/tex]
where:
* G is the universal gravitational constant
* ME is the mass of the Earth
* RE is the radius of the Earth
* omega is the angular velocity of the Earth's rotation
Substituting the given values, we get:
[tex]h = sqrt[(6.67430 x 10^-11 m^3 kg^-1 s^-2) * (6 x 10^24 kg) / ((6.4 x 10^6 m)^2 * (7.2921150 x 10^-5 rad/s)^2)] - (6.4 x 10^6 m)[/tex]
h = 35.786 km
Therefore, the height of the shuttle in geostationary orbit is 35,786 kilometers.
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Rewritten by : Barada
