The minimum energy required to launch a m kg satellite from earths surface in a circular orbit at an altitude of 2R, 'R' is the radius of earth, will be:

Question

The minimum energy required to launch a m kg satellite from earths surface in a circular orbit at an altitude of 2R, 'R' is the radius of earth, will be: (A)  3 mgR  (B)  (5/6) mgR  (C)  2 mgR  (D)  (1/5) mgR

Tardigrade Admin · Accepted Answer

The kinetic energy at altitude 2 R is =( GMm /6 R ) The gravitational potential energy at altitude 2 R is =(- GMm /3 R ) Total energy = Kinetic energy + potential energy =( GMm /6 R )+(- GMm /3 R ) Potential energy at the surfce is =( GMm / R ) Therefore, required kinetic energy =-( GMm /6 R )+( GMm / R )=(5 GMm /6 R )

Q. The minimum energy required to launch a $m k g$ satellite from earths surface in a circular orbit at an altitude of $2 R,^{'} R^{'}$ is the radius of earth, will be:

$3 m g R$

$\frac{5}{6} m g R$

$2 m g R$

$\frac{1}{5} m g R$

Q. The minimum energy required to launch a mkg satellite from earths surface in a circular orbit at an altitude of 2R,′R′ is the radius of earth, will be:

3mgR

65​mgR

2mgR

51​mgR

The kinetic energy at altitude 2R is =6RGMm​ The gravitational potential energy at altitude 2R is =3R−GMm​ Total energy = Kinetic energy + potential energy =6RGMm​+3R−GMm​ Potential energy at the surfce is =RGMm​ Therefore, required kinetic energy =−6RGMm​+RGMm​=6R5GMm​

Q. The minimum energy required to launch a $m k g$ satellite from earths surface in a circular orbit at an altitude of $2 R,^{'} R^{'}$ is the radius of earth, will be:

$3 m g R$

$\frac{5}{6} m g R$

$2 m g R$

$\frac{1}{5} m g R$