Question

planet.
3. A satellite revolves round a planet in an orbit just
above the surface of
Taking
G= 6.67 x 10-11 Nm² kg-2 and the mean density of
the planet = 8.0 x 10°kg m
= 8.0 x 103 kg m-3, find the period of
satellite.
(Ans. 4206.7 s)​

Answer 1

Answer:

Time period of the satellite ≈ 4200 s

Explanation:

Given:

• Satellite revolves around a planet just above the surface
• G = 6.67 × 10⁻¹¹ N m² kg⁻²
• Mean density of the planet = 8.0 × 10³ kg m⁻³

To Find:

The time period of the satellite

Solution:

First finding the mass of the planet.

We know that,

Mass = Density (ρ) × Volume

Hence,

Mass = 8 × 10³ × 4/3 × π R³

Mass = 10.67 × 10³ π R³

where R is the radius of the planet.

Now finding the orbital velocity of the satellite,

Since the satellite is just above the surface of earth, orbital velocity is given by,

$\boxed{\sf v_0=\sqrt{\dfrac{GM}{R}} }$

where v₀ is the orbital velocity, G is the gravitational constant, R is the radius of the planet.

Substituting the data we get,

$\sf v_0=\sqrt{\dfrac{6.67\times 10^{-11} \times 10.67\times 10^{3}\times \pi \times R^3}{R}}$

$\sf v_0=\sqrt{71.17\times 10^{-8}\times \pi \times R^2}$

$\sf v_0=R\times \sqrt{71.17\times 10^{-8}\times \pi}\: \: m/s$

$\sf v_0=R\times \sqrt{223.68\times 10^{-8}}\: \: m/s$

$\sf v_0=15\times 10^{-4}\times R \:m/s$

Now time period of a satellite is given by,

$\boxed {\sf T=\dfrac{2 \: \pi \:R}{v_0}}$

Substitute the data,

$\sf T = \dfrac{2\: \pi \: R}{R\times 15\times 10^{-4}}$

$\sf T = \dfrac{2\: \pi }{ 15\times 10^{-4} }$

$\sf T=\dfrac{6.29\times 10^{4}}{15}$

$\sf T = 0.42 \times 10^4$

$\sf T \approx 4200s$

Hence time period of the satellite is approximately 4200 s.

Answer 2

Explanation:

______________________________

$\text{\Large\underline{\red{Question:-}}}$

:$\Longrightarrow$ ● A satellite revolves round a planet in an orbit just above the surface. Taking

G= 6.67 x 10-11 Nm² kg-2 and the mean density of

the planet = 8.0 x 10°kg m = 8.0 x 103 kg m-3, find the period of satellite.

$\text{\Large\underline{\orange{To\ find:-}}}$

$\sf To\ find = \begin{cases} \sf{In\ this\ question\ we\ have\ to\ find\ the\ time\ period\ of\ the\ satellite.} \end{cases}$

$\text{\Large\underline{\green{Given:-}}}$

$\sf Given = \begin{cases} \sf{●\ Satellite\ revolves\ around\ a\ planet\ just\ above\ the\ surface.} \\ \\ \sf{●\ G\ =\ 6.67\ ×\ 10^‐¹¹\ Nm²kg^‐².} \\ \\ \sf{●\ The\ mean\ density\ of\ the\ planet\ =\ 8.0\ ×\ 10³kgm^‐³.} \end{cases}$

$\text{\Large\underline{\purple{Solution:-}}}$

We know that:-

☯️ $\sf\pink{●\ Mass\ =\ Density\ ×\ Volume.}$

Therefore:-

$\sf{●\ Mass\ =\ 8\ ×\ 10³\ ×\ \dfrac{4}{3}\ ×\ \pi\ R³.}$

$\sf{●\ Mass\ =\ 10.67\ ×\ 10³\ \pi\ R³.}$

Here, "R" is the radius of the planet.

$\star \: \sf\underbrace{Now,\ we\ will\ find\ the\ orbital\ velocity\ of\ the\ satellite} \: \star$

Orbital velocity is given by:-

$\sf{v_{0}\ =\ \sqrt \dfrac{GM}{R}}$

Here:-

● $\bf{v_{0}\ is\ the\ orbital\ velocity.}$

● $\bf{G\ is\ the\ gravitational\ constant.}$

● $\bf{R\ is\ the\ radius\ of\ the\ planet.}$

$\text{\large\underline{\blue{By\ substituting\ the\ values:-}}}$

● $\sf{v_{0}\ =\ \sqrt \dfrac{6.67×10^‐¹¹×10.67×10³×\ \pi ×R³}{R}}$

● $\sf{v_{0}\ =\ \sqrt{7.17×10^‐⁸×\ \pi ×R²}}$

● $\sf{v_{0}\ =\ R\ ×\ \sqrt{71.17×10^‐⁸×\ \pi}}$ m/s

● $\sf{v_{0}\ =\ R\ ×\ \sqrt{223.68×10^‐⁸}}$ m/s

● $\sf{v_{0}\ =\ 15\ ×\ 10^‐⁴\ ×\ R}$ m/s.

$\star \: \sf\underbrace{Now,\ time\ period\ of\ satellite} \: \star$

● $\sf\pink{T\ =\ \dfrac{2πR}{v_0}}$

By substituting the values, we get:-

● $\sf{T\ =\ \dfrac{2πR}{R×15×10^‐⁴}}$

● $\sf{T\ =\ \dfrac{2π}{15×10^‐⁴}}$

● $\sf{T\ =\ \dfrac{6.29×10⁴}{15}}$

● $\sf{T\ =\ 0.42\ ×\ 10⁴}$

● $\bf{T\ =\ \approx \ 4200s.}$

Hence:-

● The time period of the satellite is approximately 4200s.

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