Question

the binding energy of a satellite revolving around planet in a circular orbit is 3*10⁹j what is its kinetic energy​

Answer 1

Answer:

The binding energy of a satellite revolving around the planet in a circular orbit is 3×109 J. Its kinetic energy is 3 × 10+9 J.

hope it's helpful to you

Answer 2

Answer:

the answer is

$3 \times {10}^{9}J$

Explanation:

Binding energy of a satellite is the minimum amount of energy required to remove the satellite from the orbit in which it experiences gravitational attraction towards planet to infinity.

binding energy is negative energy.

Kinetic energy is due to orbital motion of satellite in it's orbit around the planet.

Let us consider M be the mass of planet and m be the mass of satellite.

$KE = \frac{GmM }{2r}$

$PE = - \frac{ GMm }{r}$

PE is potential energy.

Total energy = KE + PE

=

$- \frac{GmM }{2r}$

BE = - Total energy of satellite

BE = KE

$Binding \: Energy = Energy \: of \: the \: satellite \: at \: infinity \: - Energy \: of \: the \: satellite \: in \: orbit \\ = 0 - ( - \frac{GmM }{2r} ) \\ = \frac{GmM }{2r}$