There is a rare alignment of three planets along the x axis (perhaps this is of astrological significance). Planet 1 has mass 8 x 1024 kg and is located at x = 0 m. Planet 2 has mass 6 x 1024 kg and is located at x = 3 x 1011 m. Planet 3 has mass 15 x 1024 kg and is located at x = 27 x 1011 m. Calculate the total gravitational binding energy of this system, in units of 1026 J. Use G = 6.7 x 10-11 N m2/ kg2, and note that your answer will be positive since I said binding energy. (Please answer to the fourth decimal place - i.e 14.3225)
There is a rare alignment of three planets along the x axis (perhaps this is of astrological significance). Planet 1 has mass 8 x 1024 kg and is located at x = 0 m. Planet 2 has mass 6 x 1024 kg and is located at x = 3 x 1011 m. Planet 3 has mass 15 x 1024 kg and is located at x = 27 x 1011 m. Calculate the total gravitational binding energy of this system, in units of 1026 J. Use G = 6.7 x 10-11 N m2/ kg2, and note that your answer will be positive since I said binding energy.
(Please answer to the fourth decimal place - i.e 14.3225)
The gravitational binding energy between two masses m1 and m2 with r distance apart is given by
Where G = Gravitational constant = 6.67 × 10- 11 N.m2kg- 2
Planet 1 mass is m1 = 8 × 1024 kg
Position of planet 1 is x1 = 0 m
Planet 2 mass is m2 = 6 × 1024 kg
Position of planet 2 is x2 = 3 × 1011 m
Planet 3 mass is m3 = 15 × 1024 kg
Position of planet 3 is x3 = 27 × 1011 m
The distance between planet 1 and planet 2 is r12 = x2 - x1 = 3 × 1011 - 0 = 3 × 1011 m
The distance between planet 1 and planet 3 is r13 = x3 - x1 = 27 × 1011 - 0 = 27 × 1011 m
The distance between planet 2 and planet 3 is r23 = x3 - x2 = 27 × 1011 - 3 × 1011 = 24 × 1011 m
The total gravitational binding energy of this system is given by
In units of 1026 J
U = 161.3769
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