A satellite is in an orbit around the earth. If its kinetic energy is doubled, then

Question

A satellite is in an orbit around the earth. If its kinetic energy is doubled, then (A) it will maintain its path (B) it will fall on the earth (C) it will rotate with a great speed (D) it will escape out of earth’s gravitational field

Tardigrade Admin · Accepted Answer

Kinetic energy, K=(1/2)m upsilon02 or v0=√(2k/m) When kinetic energy is doubled, then v'0=√(2×2K/m)=√2√(2K/m)=√2v0 But √2v0=ve (escape velocity) Hence satellite will escape out of earthâs gravitational field.

Q. A satellite is in an orbit around the earth. If its kinetic energy is doubled, then

it will maintain its path

it will fall on the earth

it will rotate with a great speed

it will escape out of earth’s gravitational field

Kinetic energy, $K = \frac{1}{2} m υ_{0}^{2}$
or $v_{0} = \frac{2 k}{m}$
When kinetic energy is doubled, then
$v_{0}^{'} = \frac{2 \times 2 K}{m} = 2 \frac{2 K}{m} = 2 v_{0}$
But $2 v_{0} = v_{e}$ (escape velocity)
Hence satellite will escape out of earth’s gravitational field.

Q. A satellite is in an orbit around the earth. If its kinetic energy is doubled, then

it will maintain its path

it will fall on the earth

it will rotate with a great speed

it will escape out of earth’s gravitational field

Kinetic energy, K=21​mυ02​ orv0​=m2k​​ When kinetic energy is doubled, then v0′​=m2×2K​​=2​m2K​​=2v0​​ But 2v0​​=ve​ (escape velocity) Hence satellite will escape out of earth’s gravitational field.

Kinetic energy, $K = \frac{1}{2} m υ_{0}^{2}$
or $v_{0} = \frac{2 k}{m}$
When kinetic energy is doubled, then
$v_{0}^{'} = \frac{2 \times 2 K}{m} = 2 \frac{2 K}{m} = 2 v_{0}$
But $2 v_{0} = v_{e}$ (escape velocity)
Hence satellite will escape out of earth’s gravitational field.