Chapter 7: Gravitational Field Flashcards
Explain how a man feels “weightlessness” in space orbit
Where the gravitational force acting on the man is just sufficient to provide
the centripetal force required to keep him in orbit.
such that there is no contact force acting on the man.
Hence, man feels weightless.
State Newton’s Law of Gravitation
Two point masses will attract each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them
F = (GMm) / r²
Define a gravitational field
A region of space where a mass experiences a force due to the presence of another mass
Define gravitational field strength (g)
The gravitational force per unit mass acting on a test mass placed at that point
Define a geostationary orbit
A circular orbit in which a body appears to be stationary to a person on the planet’s surface (geo= Earth)
Define gravitational potential (Φ)
The work done per unit mass by an external agent in bringing a test mass from infinity to that point without a change in kinetic energy
Φ = - (GM) / r
Define gravitational potential energy (U)
The work done by an external agent in bringing a mass (m) from infinity to that point without a change in kinetic energy
U = - (GMm) / r
State the relationship between gravitational force (Fg) and gravitational potential energy (U)
Fg = - dU / dr
(Gravitational force is the negative gradient of gravitational potential energy)
State the relationship between gravitational field strength (g) and gravitational potential (Φ)
g = - dΦ / dr
(Gravitational field strength is the negative gradient of gravitational potential)
Explain how escape velocity is determined
By C.O.E. , total initial energy = total final energy
(Initial KE) + (Initial PE) = (Final KE) + (Final PE)
½mv² + (- GMm / R) = 0
v = √(2GM)/R
Express kinetic energy in terms of G, M, m and r
(for a body of mass m orbiting at distance r from the centre of a planet of mass M)
Subbing [v = √(2GM)/R] into [KE = ½mv²],
KE = GMm / 2r
Explain why, near the surface of the Earth, the gravitational field strength is equal to the acceleration of free fall.
Explain why g field strength has same mag and direction as the centripetal a of the moon
The gravitational force of Jupiter acting on each its moons provides the centripetal force for their circular motions.
By Newton’s 2nd Law of Motion, this gravitational force causes the rate of change of momentum of the moon (mmoon ac), with the change in momentum taking place in the direction of this gravitational force.
Since gravitational field strength is the gravitational force per unit mass, it will hence be numerically equal to the centripetal acceleration (ac) of the moon, with the direction of the centripetal acceleration in the direction of the gravitational field strength.
