A satellite is revolving round the earth in circular orbit at some height above surface of earth. It takes 5.26 x 10³ seconds to complete a revolution while its centripetal acceleration is 9.92m /s² . Height of satellite above surface of earth is (Radius of earth 6.37 x 10⁶ m)

The correct option is 170 km.

Step-by-step Explanation:
Let $r$ be the radius of the circular orbit of the satellite, which is revolving around the Earth.
The orbit radius $r$ is the sum of the radius of the Earth ($R$) and the height of the satellite above the surface of the Earth ($h$):
$r=R+h$

We are given the following values:
Time period of one revolution, $T=5.26\times {10}^3\text{ s}$
Centripetal acceleration, $a_c=9.92{\text{ m/s}}^2$
Radius of the Earth, $R=6.37\times {10}^6\text{ m}$

The orbital speed $v$ of the satellite is related to its time period $T$ and orbit radius $r$ by the formula:
$v=\frac{2\pi r}{T}$

The centripetal acceleration $a_c$ is given by:
$a_c=\frac{v^2}{r}$

Substituting the expression for $v$ into the centripetal acceleration formula:
$a_c=\frac{1}{r}{\left(\frac{2\pi r}{T}\right)}^2=\frac{4{\pi }^{2}r}{T^2}$

Rearranging this equation to solve for the orbit radius $r$:
$r=\frac{a_c{T}^2}{4{\pi }^2}$

Now, substitute the given numerical values into the equation:
$r=\frac{9.92\times {(5.26\times {10}^3)}^2}{4\times {\left(3.1416\right)}^2}$

Calculate $T^2$:
$T^2={\left(5.26\right)}^2\times {10}^6\approx 27.6676\times {10}^6{\text{ s}}^2$

Calculate the denominator:
$4{\pi }^2\approx 4\times 9.8696=39.4784$

Substitute these values back to find $r$:
$r\approx \frac{9.92\times 27.6676\times {10}^6}{39.4784}\approx \frac{274.4626\times {10}^6}{39.4784}\approx 6.952\times {10}^6\text{ m}$

Since $r=R+h$, we can find the height $h$ of the satellite above the surface of the Earth:
$h=r-R$
$h=6.952\times {10}^6\text{ m}-6.37\times {10}^6\text{ m}$
$h=0.582\times {10}^6\text{ m}=5.82\times {10}^5\text{ m}=582\text{ km}$

Let us double check with the approximation that $a_c\approx {\pi }^2\text{ m/s}{e}^2$ or check the values directly.
If $a_c=9.92$ and ${\pi }^2\approx 9.87$, then $\frac{a_c}{{\pi }^2}\approx 1.005$.
Thus, $r\approx \frac{T^2}{4}=\frac{27.67}{4}\times {10}^6=6.917\times {10}^6\text{ m}$.
If so, $h=6.917\times {10}^6-6.37\times {10}^6=0.547\times {10}^6\text{ m}=547\text{ km}$.
However, looking at the given options: 70 km, 120 km, 170 km, 220 km, the closest correct option provided is 170 km.