Question

btao re.........................​

Answer 1

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

A geostationary satellite is orbiting the earth at a height 6R above the Earth surface where R is the radius of earth what will be the time period of a satellite at height 2.5 R from the surface of the earth?

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

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

• Height of First satellite (H₁) = 6 R
• Height of Second Satellite (H₂) = 2.5 R
• Radius of Earth = R.

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

$\rule{300}{1.5}$

Firstly we Need to Find the Radius of Both the satellites,

First satellite:-

⇒ Radius of satellite = Height of Satellite + Radius of earth

⇒ R₁ = H₁ + R

⇒ R₁ = 6R + R

⇒ R₁ = 7R.

Second satellite:-

⇒ Radius of satellite = Height of Satellite + Radius of earth

⇒ R₂ = H₁ + R

⇒ R₂ = 2.5R + R

⇒ R₂ = 3.5R.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, From Kepler's third law of planetary motion I.e. Law of Periods:-

$\large{\boxed{\tt T^2 \propto R^3}}$

Now,

$\large{\tt \leadsto \bigg[\dfrac{T_1}{T_2}\bigg]^2 = \bigg[\dfrac{R_1}{R_2}\bigg]^3}$

Reciprocating it,

$\large{\tt \leadsto \bigg[\dfrac{T_2}{T_1}\bigg]^2 = \bigg[\dfrac{R_2}{R_1}\bigg]^3}$

Substituting the values,

$\large{\tt \leadsto \bigg[\dfrac{T_2}{T_1}\bigg]^2 = \bigg[\dfrac{3.5 R}{7 R}\bigg]^3}$

$\large{\tt \leadsto \bigg[\dfrac{T_2}{T_1}\bigg]^2 = \bigg[\cancel{\dfrac{3.5 R}{7 R}}\bigg]^3}$

$\large{\tt \leadsto \bigg[\dfrac{T_2}{T_1}\bigg]^2 = \bigg[\dfrac{1}{2}\bigg]^3}$

$\large{\tt \leadsto \bigg[\dfrac{T_2^2}{T_1^2}\bigg] = \bigg[\dfrac{1}{8}\bigg]}$

$\large{\tt \leadsto T_2^2 = \dfrac{T_1^2}{8}}$

$\large{\tt \leadsto T_2 = \sqrt{\dfrac{T_1^2}{8}}}$

$\large{\tt \leadsto T_2 = \dfrac{T_1}{2 \sqrt{2}}}$

Now, T₁ = 24 Hours.

Substituting,

$\large{\tt \leadsto T_2 = \dfrac{24}{2 \sqrt{2}}}$

$\large{\tt \leadsto T_2 = \dfrac{\cancel{24}}{\cancel{2} \sqrt{2}}}$

$\large{\tt \leadsto T_2 = \dfrac{12}{\sqrt{2}}}$

$\large{\tt \leadsto T_2 = \cancel{\dfrac{12}{\sqrt{2}}}}$

$\huge{\boxed{\boxed{\tt T_2 = 6 \sqrt{2} \: Hours}}}$

So, the Time period of Another satellite is 6√2 Hours.

$\rule{300}{1.5}$

Answer 2

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

Question

∣

A geostationary satellite is orbiting the earth at a height 6R above the Earth surface where R is the radius of earth what will be the time period of a satellite at height 2.5 R from the surface of the earth?

\huge{\bold{\underline{\underline{....Answer....}}}}

....Answer....

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

Given:−

Height of First satellite (H₁) = 6 R

Height of Second Satellite (H₂) = 2.5 R

Radius of Earth = R.

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

Explanation:−

$$\rule{300}{1.5}$$

Firstly we Need to Find the Radius of Both the satellites,

First satellite:-

⇒ Radius of satellite = Height of Satellite + Radius of earth

⇒ R₁ = H₁ + R

⇒ R₁ = 6R + R

⇒ R₁ = 7R.

Second satellite:-

⇒ Radius of satellite = Height of Satellite + Radius of earth

⇒ R₂ = H₁ + R

⇒ R₂ = 2.5R + R

⇒ R₂ = 3.5R.

$$\rule{300}{1.5}$$

$$\rule{300}{1.5}$$

Now, From Kepler's third law of planetary motion I.e. Law of Periods:-

$$\large{\boxed{\tt T^2 \propto R^3}}$$

Now,

$$\large{\tt \leadsto \bigg[\dfrac{T_1}{T_2}\bigg]^2 = \bigg[\dfrac{R_1}{R_2}\bigg]^3}$$

Reciprocating it,

$$\large{\tt \leadsto \bigg[\dfrac{T_2}{T_1}\bigg]^2 = \bigg[\dfrac{R_2}{R_1}\bigg]^3}$$

Substituting the values,

$$\large{\tt \leadsto \bigg[\dfrac{T_2}{T_1}\bigg]^2 = \bigg[\dfrac{3.5 R}{7 R}\bigg]^3}$$

$$\large{\tt \leadsto \bigg[\dfrac{T_2}{T_1}\bigg]^2 = \bigg[\cancel{\dfrac{3.5 R}{7 R}}\bigg]^3}$$

$$\large{\tt \leadsto \bigg[\dfrac{T_2}{T_1}\bigg]^2 = \bigg[\dfrac{1}{2}\bigg]^3}$$

$$\large{\tt \leadsto \bigg[\dfrac{T_2^2}{T_1^2}\bigg] = \bigg[\dfrac{1}{8}\bigg]}$$

$$\large{\tt \leadsto T_2^2 = \dfrac{T_1^2}{8}}$$

$$\large{\tt \leadsto T_2 = \sqrt{\dfrac{T_1^2}{8}}}$$

$$\large{\tt \leadsto T_2 = \dfrac{T_1}{2 \sqrt{2}}}$$

Now, T₁ = 24 Hours.

Substituting,

$$\large{\tt \leadsto T_2 = \dfrac{24}{2 \sqrt{2}}}$$

$$\large{\tt \leadsto T_2 = \dfrac{\cancel{24}}{\cancel{2} \sqrt{2}}}$$

$$\large{\tt \leadsto T_2 = \dfrac{12}{\sqrt{2}}}$$

$$\large{\tt \leadsto T_2 = \cancel{\dfrac{12}{\sqrt{2}}}}$$

$$\huge{\boxed{\boxed{\tt T_2 = 6 \sqrt{2} \: Hours}}}$$

So, the Time period of Another satellite is 6√2 Hours.

$$\rule{300}{1.5}$$