Gravitational force between point masses
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Explain why we can treat a uniform sphere's mass as a point mass concentrated at its center when calculating gravitational force outside the sphere.
Gravitational field lines outside a uniform sphere are identical to those of a point mass at its center. This simplifies gravitational force calculations using Newton's Law of Gravitation, allowing us to treat the sphere as if all its mass is at a single point.
State Newton's Law of Gravitation.
Newton's Law of Gravitation states that the gravitational force (F) between two point masses (m1 and m2) is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between their centers: F = Gm1m2 / r²
A satellite orbits a planet. How is the gravitational force related to the centripetal acceleration required for its orbit?
The gravitational force between the planet and the satellite provides the centripetal force necessary for the satellite's circular motion. Therefore, F_gravitational = F_centripetal, which means Gm1m2/r^2 = mv^2/r, where m is the satellite's mass.
What are the defining characteristics of a geostationary orbit?
A geostationary orbit is a circular orbit where the satellite remains at a fixed point above the Earth's surface. This requires an orbital period of 24 hours, orbiting from west to east, directly above the Equator.
How would you calculate the orbital speed (v) of a satellite orbiting a planet of mass M at a distance r from its center?
Equate gravitational force to centripetal force (GMm/r^2 = mv^2/r), and solve for v. Therefore, v = sqrt(GM/r), where G is the gravitational constant, M is the planet's mass, and r is the orbital radius.
If the orbital radius of a satellite is doubled, how does this affect its orbital period?
From Kepler's Third Law (T² ∝ r³), if the orbital radius is doubled (r' = 2r), then the square of the new period (T'²) will be proportional to (2r)³. This means T'² = 8T², so the new period T' is sqrt(8) times the original period T.
A satellite orbits a planet at a distance 'r' with a velocity 'v'. If the distance is increased to '4r', what will be the new orbital velocity?
Since v = sqrt(GM/r), the velocity is inversely proportional to the square root of the radius. If the radius increases to 4r, the new velocity v' = sqrt(GM/4r) = v/2. The new velocity will be half of the original velocity.
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