Question

Consider a spacecraft that is far away from planets or other massive objects. The mass of the spacecraft is M = 1.5×105 kg. The rocket engines are shut off and the spacecraft coasts with a velocity vector v = (0, 20, 0) km/s. The space craft passes the position x = (12, 15, 0) km at which time the spacecraft fires its thruster rockets giving it a net force of F = (6 × 104 , 0, 0) N which is exerted for 3.4 s. The ejected gases have total mass that is small compared to the mass of the spacecraft. a) Where is the space craft 1 hour afterwards? b) What approximations have you made in your analysis?

Kepler’s second law is this statement: A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. We are going to prove this statement. Consider the wedge in the figure with area dA = 1 2 R 2 dθ The rate that area is swept per unit time is dA dt = 1 2 R 2 dθ dt = 1 2 R 2 ˙θ and this is true even if radius R is varying. We take the origin to be the center of the Sun and radius R is the distance between planet and Sun. The angle θ gives the position of the planet in the ecliptic plane. Kepler’s second law is equivalent to dA dt = constant or d 2A dt = 0. In class we showed that acceleration in polar coordinates can be written a = (R¨ − R ˙θ 2 )ˆr + (2R˙ ˙θ + R¨θ)θˆ Because the gravitational force is in the radial direction, the tangential component of acceleration is zero. This means that 2R˙ ˙θ + R¨θ = 0 Show that this relation is equivalent to dA/dt = constant and Kepler’s second law.

A spherical hollow is made in a sphere of radius R = 11.3 cm such that its surface touches the outside surface of the sphere and passes through its center (see Figure). The mass of the sphere before hollowing was M = 57.0 kg. What is the magnitude of the gravitational force between the hollowed-out lead sphere and a small sphere of mass m = 4.2 kg, located a distance d = 0.55 m from the center of the lead sphere?

Answer 1

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