The distance between centre of the earth and moon is 384000 km. If the mass of the earth is 6 x 10²⁴ kg and G = 6.67 x 10⁻¹¹ Nm² /kg². The speed of the moon is nearly

The correct option is 1 km / sec.

To find the orbital speed of the moon around the Earth, we use the principle that the gravitational force of attraction between the Earth and the moon provides the necessary centripetal force for the moon's circular motion.

The centripetal force ($F_c$) acting on the moon is given by:
$F_c=\frac{m{v}^2}{r}$
where:
• $m$ is the mass of the moon,
• $v$ is the orbital speed of the moon, and
• $r$ is the distance between the center of the Earth and the moon.

The gravitational force ($F_g$) between the Earth and the moon is given by Newton's law of gravitation:
$F_g=\frac{GMm}{r^2}$
where:
• $M$ is the mass of the Earth, and
• $G$ is the universal gravitational constant.

Equating the centripetal force and the gravitational force:
$\frac{m{v}^2}{r}=\frac{GMm}{r^2}$

We can simplify this equation by cancelling the mass of the moon ($m$) and one factor of $r$ from both sides:
$v^2=\frac{GM}{r}$

Taking the square root of both sides gives the expression for the speed of the moon:
$v=\sqrt{\frac{GM}{r}}$

Now, let's identify the given values from the problem statement:
• Mass of the Earth, $M=6\times {10}^{24}\text{kg}$
• Distance, $r=384000\text{km}=3.84\times {10}^8\text{m}$
• Gravitational constant, $G=6.67\times {10}^{-11}{\text{N m}}^2/{\text{kg}}^2$

Substituting these values into the formula:
$v=\sqrt{\frac{(6.67\times {10}^{-11})\times (6\times {10}^{24})}{3.84\times {10}^8}}$

Simplify the numerator:
$6.67\times 6=40.02$
${10}^{-11}\times {10}^{24}={10}^{13}$
So, the numerator is $40.02\times {10}^{13}$.

Now, divide by the denominator:
$v=\sqrt{\frac{40.02\times {10}^{13}}{3.84\times {10}^8}}$
$v=\sqrt{10.422\times {10}^5}$
$v=\sqrt{1.0422\times {10}^6}$
$v\approx 1.02\times {10}^3\text{m/s}$

Convert the speed into kilometers per second (km/sec):
$v\approx 1.02\text{km/sec}$

Therefore, the speed of the moon is nearly 1 km / sec.