Question

Taking the radius of the earth to be 6400 km, by what percentage will the acceleration due to
at a height of 100 km from the surface of the earth differ from that on the surface of the earth
a) About 1.5% b) About 5% c) About 8% d)About 3%​

Answer 1

Answer:

Given:

Object is taken to a height of 100 km.

To find:

Percentage decrease in gravitational acceleration.

Formulas used:

At a height "h" from the Earth surface ;

$g = G \: \frac{m}{ {(r + h) \: }^{2} }$

where m is the mass of Earth, r is the radius of the Earth and h is the height .

If r >>> h , then we can apply approximations such as

$g" = g ( \: 1 - \frac{2h}{r} ) \:$

Calculation:

g" = g { 1 - (2 ×100)/6400}

=> g" = g ( 1 - 1/32)

=> g" = (31/32) g

So change in gravity ∆g = g - g"

=> ∆g = g - (31/32)g

=> ∆g = g/32.

% change = ∆g/g × 100%

=> % change = 1/32 × 100%

=> % change = 3.125 % ≡ 3%.

So final answer is about 3% decrease ( option d).

Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

Taking the radius of the earth to be 6400 km, by what percentage will the acceleration due to at a height of 100 km from the surface of the earth differ from that on the surface of the earth?

$\huge{\bold{\underline{\underline{Answer}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Radius of Earth (R) = 6400 Km
• Height to which the Body is taken (h) = 100 km.
• Let the Acceleration due to gravity at surface of earth be "g".

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

From the relation between Acceleration due to gravity (g) & Height (h),

$\large{\boxed{\tt g_h = g\left[\dfrac{R}{R + h}\right]^2}}$

$\large{\tt \hookrightarrow \dfrac{g_h}{g} = \left[\dfrac{R}{R + h}\right]^2}$

But Here the Given Height (h) is very small compared to the Radius of the Earth.

[h << R]

Therefore, the Formula Becomes,

$\large{\tt \hookrightarrow \dfrac{g_h}{g} = \left[1 - \dfrac{2h}{R}\right]}$

$\large{\boxed{\tt g_h = g \left[1 - \dfrac{2h}{R}\right]}}$

Substituting the values,

$\large{\tt \hookrightarrow g_h = g \left[ 1 - \dfrac{2 \times 100}{6400} \right]}$

$\large{\tt \hookrightarrow g_h = g \left[ 1 - \dfrac{200}{6400} \right]}$

$\large{\tt \hookrightarrow g_h = g \left[ 1 - \cancel{\dfrac{200}{6400}} \right]}$

$\large{\tt \hookrightarrow g_h = g \left[ 1 - \dfrac{1}{32} \right]}$

$\large{\tt \hookrightarrow g_h = g \left[ \dfrac{32 - 1}{32} \right]}$

$\large{\tt \hookrightarrow g_h = g \left[ \dfrac{31}{32} \right]}$

$\large{\hookrightarrow{\underline{\underline{\tt g_h = \dfrac{31}{32}g}}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now,

Finding the Difference in percentage of Acceleration due to gravity will differ.

$\large{\boxed{\tt \Delta g \% = \dfrac{g - g_h}{g} \times 100}}$

Substituting the values,

$\large{\tt \hookrightarrow \Delta g \% = \dfrac{g - \dfrac{31}{32}g}{g} \times 100}$

$\large{\tt \hookrightarrow \Delta g \% = \dfrac{\dfrac{32g -31g}{32}}{g} \times 100}$

$\large{\tt \hookrightarrow \Delta g \% = \dfrac{\dfrac{1g}{32}}{g} \times 100}$

$\large{\tt \hookrightarrow \Delta g \% = \dfrac{\dfrac{\cancel{1g}}{32}}{\cancel{g}} \times 100}$

$\large{\tt \hookrightarrow \Delta g \% = \dfrac{1}{32} \times 100}$

$\large{\tt \hookrightarrow \Delta g \% = \dfrac{1}{\cancel{32}} \times \cancel{100}}$

$\large{\tt \hookrightarrow \Delta g \% = 3.125}$

$\huge{\boxed{\boxed{\tt \Delta g \% \approx 3 \%}}}$

So, Acceleration will Differ by About 3% From the surface of earth (Option- d)

$\rule{300}{1.5}$