imported>Patrick: Created page with 'The distance between the Earth and the Sun is $1.5 \times 10^{11} m$ and the masses of these objects are $6 \times 10^{24} kg$ and $2 \times 10^{30} kg…'$

2011-09-29T22:12:16Z

Created page with 'The distance between the Earth and the Sun is <math>1.5 \times 10^{11} m</math> and the masses of these objects are <math>6 \times 10^{24} kg</math> and <math>2 \times 10^{30} kg…'

New page

The distance between the Earth and the Sun is <math>1.5 \times 10^{11} m</math> and the masses of these objects are <math>6 \times 10^{24} kg</math> and <math>2 \times 10^{30} kg</math>. The Earth is orbiting the Sun and therefore the energy of the system is

<math>E = \frac{1}{2} U = -\frac{1}{2} {G M_S M_E \over r} = -\frac{1}{2} {(6.67 \times 10^{-11})(2 \times 10^{30})(6 \times 10^{24}) \over 1.5 \times 10^{11}} = -5.3 \times 10^{33} J</math>
<br><br>
The escape velocity at the distance <math>r = 1.5 \times 10^{11} m</math> is given by

<math>v = \sqrt{2 G M_S \over r} = \sqrt{2 (6.67 \times 10^{-11}) (2 \times 10^{30}) \over 1.5 \times 10^{11}} = 4.22 \times 10^4 m/s = 42.2 km/s</math>

Satellites and Binding Energy EX 10 - Revision history

imported>Patrick: Created page with 'The distance between the Earth and the Sun is 1.5 \times 10^{11} m and the masses of these objects are 6 \times 10^{24} kg and 2 \times 10^{30} kg…'

imported>Patrick: Created page with 'The distance between the Earth and the Sun is $1.5 \times 10^{11} m$ and the masses of these objects are $6 \times 10^{24} kg$ and $2 \times 10^{30} kg…'$