The value of g on the earth's surface is 980 cm/sec² . Its value at a height of 64 km from the earth's surface is (Radius of the earth R = 6400 Kilometers)

The correct answer is 960.40cm / sec².

Step-by-step Explanation:

The acceleration due to gravity at a height $h$ above the surface of the Earth is given by the formula:
$g^{\prime }=g{\left(\frac{R}{R+h}\right)}^2$
where:
• $g$ is the acceleration due to gravity on the Earth's surface ($980\text{cm/sec²}$).
• $R$ is the radius of the Earth ($6400\text{km}$).
• $h$ is the height above the Earth's surface ($64\text{km}$).

Since the height $h$ ($64\text{km}$) is very small compared to the radius of the Earth $R$ ($6400\text{km}$), we can use the binomial approximation:
$g^{\prime }\approx g\left(1-\frac{2h}{R}\right)$

Now, let's substitute the given values into the approximation formula:

$g^{\prime }=980\times \left(1-\frac{2\times 64}{6400}\right)$

Simplify the fraction inside the parentheses:
$\frac{2\times 64}{6400}=\frac{128}{6400}=0.02$

Substitute this back into the equation:
$g^{\prime }=980\times (1-0.02)$
$g^{\prime }=980\times 0.98$
$g^{\prime }=960.40\text{cm/sec²}$

Therefore, the value of gravity at a height of 64 km is 960.40cm / sec².