1. Speed of the satellite in circular orbit

For a satellite moving in a circular orbit of radius $r$ around Earth,

$v^2=\frac{GM}{r}$

where $M$ is the mass of Earth.

So the satellite’s orbital speed is $v$.

2. Condition for just escaping Earth’s gravitational field

If the object is ejected so that it just escapes, then its total mechanical energy after ejection must be zero.

Thus,

$K + U = 0$

At distance $r$ from Earth’s center,

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

So,

$K = \frac{GMm}{r}$

Using

$v^2=\frac{GM}{r}$

we get

$K = m v^2$

3. Required kinetic energy of the object at ejection

Hence, at the instant of ejection, the object must have kinetic energy

$\boxed{mv^2}$

4. Check options

• A: $mv^2$ ✅
• B: $\frac12 mv^2$ ❌
• C: $\frac32 mv^2$ ❌
• D: $2mv^2$ ❌

So the correct option is A.

5. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

They agree.