(10%) Problem 5: Planet 1 has mass 3M and radius R, while Planet 2 has mass 4M a
ID: 2038865 • Letter: #
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(10%) Problem 5: Planet 1 has mass 3M and radius R, while Planet 2 has mass 4M and radius 2R. They are separated by center-to-center distance 8R. A rock is placed halfway between their centers at point O. It is released from rest, and you may ignore any motion of the planets. -- 8R - 2R -+ * 3M 3M Otheexpertta.com 25% Part (a) Derive an expression, in terms of relevant system parameters, for the magnitude of the acceleration a of the rock at the moment it is released. > * 25% Part (b) Calculate the magnitude of the acceleration of the rock (in m/s2) the moment it is released, using M = 53 x 1022 kg and R = 39 x 100 km. Grade Summary lal = 1.45 * 10-71 100% 0 % Deductions Potential 1 sin() cos() tan() cotan asin acos() atan acotan sinh() cosh) | tanh) | cotanh). O Degrees Radians ( 7 8 9 MA 4 5 6 7 * 1 2 3 HOME - ? END CLEAR Submissions Attempts remaining: 0% per attempt) detailed view 0 % VO BACKSPACE DEL Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedback. B A 25% Part (0) The - 4 25% Part (c) The rock is released from rest at point 0. Derive an expression for the speed v with which the rock crashes into a planet. A 25% Part (d) Calculate the speed the rock crashes into the planet in m/s. case from the data in prijeten e presion for the spExplanation / Answer
(A) Fg = G m1 m2 / d^2
F_rock = G(4M)(m_rock)/(4R)^2 - G(3M)(m_rock) / (4 R)^2
F_rock = (G M m_rock) / (16 R^2)
a = F / m = G M / (16 R^2)
(B) a = (6.67 x 10^-11) (53 x 10^22) / (16 (39 x 10^8 x 10^3)^2)
a = 1.45 x 10^-13 m/s^2
(C) PEi + KEi = PEf + KEF
- (G (4 M)( m) / 4 R ) - (G (3 M) m / 4 R) + 0 = -(G(4M) m / 2 R) - (G (3M) m / 6 R ) + m v^2 / 2
- (G M / R ) (4/4 + 3/4 - 4/2 - 3/6) = v^2 / 2
3 G M / 4 R = v^2 / 2
v = sqrt[ 3 G M / 2 R ]
(D) v = sqrt[ (3 x 6.67 x 10^-11 x 53 x 10^22) /(2 x 39 x 10^8 x 10^3)]
v = 3.69 m/s
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