A satellite is orbiting close to the earth and has a kinetic energy K. The minimum extra kinetic energy required by it to just overcome the gravitation pull of the earth is

Question

A satellite is orbiting close to the earth and has a kinetic energy K. The minimum extra kinetic energy required by it to just overcome the gravitation pull of the earth is (A) √3K (B) K (C) 2√2K (D) 2K

Tardigrade Admin · Accepted Answer

We know ve=√2 v0 When satellite is in orbit K=(1/2) mv02 For satellite to escape its velocity should be ve ∴ K.E. To escape, Ke=(1/2) m ve2=(1/2) m (√2v0)2=2k ∴ Extra K.E. required=2K-K=K

Q. A satellite is orbiting close to the earth and has a kinetic energy K. The minimum extra kinetic energy required by it to just overcome the gravitation pull of the earth is

$3 K$

K

$22 K$

2K

We know $v_{e} = 2 v_{0}$
When satellite is in orbit $K = \frac{1}{2} m v_{0}^{2}$
For satellite to escape its velocity should be $v_{e}$
$∴$ K.E. To escape, $K_{e} = \frac{1}{2} m v_{e}^{2} = \frac{1}{2} m (2 v_{0})^{2} = 2 k$
$∴$ Extra K.E. required=2K-K=K

Q. A satellite is orbiting close to the earth and has a kinetic energy K. The minimum extra kinetic energy required by it to just overcome the gravitation pull of the earth is

3​K

K

22​K

2K

We know ve​=2​v0​ When satellite is in orbit K=21​mv02​ For satellite to escape its velocity should be ve​ ∴ K.E. To escape, Ke​=21​mve2​=21​m(2​v0​)2=2k ∴ Extra K.E. required=2K-K=K

$3 K$

$22 K$

We know $v_{e} = 2 v_{0}$
When satellite is in orbit $K = \frac{1}{2} m v_{0}^{2}$
For satellite to escape its velocity should be $v_{e}$
$∴$ K.E. To escape, $K_{e} = \frac{1}{2} m v_{e}^{2} = \frac{1}{2} m (2 v_{0})^{2} = 2 k$
$∴$ Extra K.E. required=2K-K=K