Physics Gravitation help...?
hello People
i need some help with my physics revision sheet. i cant seem to get my head round it. if anyone could help by telling me how to work it out to get the right answer i would be grateful...
answers would also be appreciated.
ok here is the question:
mass of the earth = 5.97 * 10^24 kg
Polar Radius of Earth = 6.36 * 10^6 m
1) calculate the net force on a 1.08 * 10^3 Kg Satellite when it is in a polar orbit 2.02 * 10^7 m above the earths orbit....
2)show that the only stable orbit for the satellite orbiting at an altitude of 2.02 * 10^7 m has a period of appoximatly 12 hours.
3) for any object orbting around a primary body R^3 �å T^2
where R is the radius of the orbit and T is the time period for the orbit.
show that thi is true and in doing so:
- state the conditions required for a stable orbit
- show that the conditions do not depend on the mass of the orbiting object.
4) discuss the particular requirements for an orbit that will keep the a satellite vertically above a certain point on earths surface.
Sure, I can help you with your physics questions on gravitation. Let's go through each question step by step.
1) To calculate the net force on a satellite in a polar orbit, you can use the formula for gravitational force:
F = G * (m1 * m2) / r^2
Where F is the net force, G is the gravitational constant (approximately 6.67 * 10^-11 Nm^2/kg^2), m1 is the mass of the satellite, m2 is the mass of the Earth, and r is the distance between the satellite and the center of the Earth.
In this case, m1 = 1.08 * 10^3 kg, m2 = 5.97 * 10^24 kg, and r = 2.02 * 10^7 m + 6.36 * 10^6 m (since the satellite is above the Earth's radius).
Substituting these values into the formula, you can calculate the net force on the satellite.
2) To show that the only stable orbit for the satellite has an approximately 12-hour period, you can use Kepler's third law of planetary motion, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit:
T^2 = (4π^2 / GM) * r^3
Where T is the orbital period, G is the gravitational constant, M is the mass of the Earth, and r is the distance between the satellite and the center of the Earth.
In this case, r = 2.02 * 10^7 m + 6.36 * 10^6 m, and you can substitute these values into the formula to solve for T. If you find that T is approximately 12 hours, it confirms that the orbit is stable.
3) The equation R^3 = T^2 relates the radius of an orbit (R) and the time period of the orbit (T). This equation is derived from Kepler's third law, and it holds true for any object orbiting a primary body.
To show that this equation is true, you can use the gravitational force equation (F = G * (m1 * m2) / r^2) and Newton's second law (F = m1 * a) to derive the equation R^3 = T^2.
By manipulating these equations, you will find that R^3 = (GM * T^2) / (4π^2), which is the same as the given equation. This shows that the equation holds true.
The conditions required for a stable orbit are that the net force on the object must be balanced by the centripetal force of the orbit. This happens when the gravitational force between the two objects (the primary body and the orbiting object) is equal to the centripetal force. As long as this balance is maintained, the object will continue to orbit.
It's important to note that the conditions for a stable orbit do not depend on the mass of the orbiting object. This means that the mass of the satellite does not affect the stability of its orbit. Only the mass of the primary body (in this case, the Earth) and the radius of the orbit determine the stability.
4) To keep a satellite vertically above a certain point on the Earth's surface, it needs to be in a geostationary orbit. This means that the satellite's orbital period matches the rotation period of the Earth, approximately 24 hours.
In a geostationary orbit, the satellite stays above the same point on the Earth's equator at all times. To achieve this, the satellite must orbit at an altitude of approximately 35,786 kilometers (22,236 miles).
This specific altitude and the matching orbital period allow the satellite to remain fixed relative to the Earth's surface, which is crucial for communication and observation purposes.
I hope this helps! Let me know if you have any further questions.