Question

A skylab of mass m kg is first launched from the surface of the earth in a circular orbit of radius 2R (from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius 3R. The minimum energy required to place the lab in the first orbit and to shift the lab from first orbit to the second orbit are

Answer 1

Answer:

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Answer 2

Explanation:

$</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1$