Respuesta :
The radius and time period of orbit of a sattellite are related by the formula:
R³ = GMT²/4π²; where R is the orbital radius, T is the orbital time period, M is the mass of the more massive body, G is the graviational constant
(9.4 x 10⁶)³ = (6.674 × 10⁻¹¹)(M)(2.8 x 10⁴)² / (4π²)
M = 6.27 x 10²³ kg
R³ = GMT²/4π²; where R is the orbital radius, T is the orbital time period, M is the mass of the more massive body, G is the graviational constant
(9.4 x 10⁶)³ = (6.674 × 10⁻¹¹)(M)(2.8 x 10⁴)² / (4π²)
M = 6.27 x 10²³ kg
The mass of the planet Mars is [tex]\boxed{6.27\times{{10}^{23}}\,{\text{kg}}}[/tex] .
Further Explanation:
Given:
The time period of rotation of the satellite is [tex]2.8 \times{10^4}\,{\text{s}}[/tex] .
The orbital radius of the satellite about the planet is [tex]9.4\times{10^6}\,{\text{m}}[/tex] .
Concept:
The relation between the mass of the planet, time period of rotation of its satellite and the orbital radius of the satellite is given by the Kepler’s third Law of planetary motion. According to this law, the square of time period of the satellite is directly proportional to the cube of the radius of the circular path of the satellite.
The Kepler’s third law is expressed mathematically as:
[tex]{T^2}=\left({\frac{{4{\pi^2}}}{{GM}}}\right){R^3}[/tex]
Here, [tex]T[/tex] is the time period of rotation of the satellite for the planet, [tex]G[/tex] is the gravitational constant, [tex]M[/tex] is the mass of the planet and [tex]R[/tex] is the orbital radius of the satellite.
The value of the gravitational constant [tex]G[/tex] is [tex]6.67\time {10^{-11}}\,{{{\text{N}}\cdot{{\text{m}}^{\text{2}}}}\mathord{\left/{\vphantom{{{\text{N}}\cdot{{\text{m}}^{\text{2}}}}{{\text{k}}{{\text{g}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{\text{k}}{{\text{g}}^{\text{2}}}}}[/tex]
Substitute the values of [tex]T[/tex] , [tex]G[/tex] and [tex]R[/tex] in above expression.
[tex]\begin{aligned}\left(2.8\times10^{4}\right)^{2}&=\left(\dfrac{4\times(3.14)^{2}}{(6.767\times10^{-11})M}\right)(9.4\times10^{6})^{3}\\M&=\dfrac{4\times(3.14)^{2}\times(9.4\times10^{6})^{3}}{(6.67\times10^{-11})(2.8\times10^{4})^{2}}\\&=\dfrac{3.279\times10^{22}}{5.229\times10^{-2}}\\&=6.27\times10^{23}\text{ Kg}\end{aligned}[/tex]
Thus, the mass of the planet Mars is [tex]\boxed{6.27\times{{10}^{23}}\,{\text{kg}}}[/tex]
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Gravitation
Keywords:
Mars, satellite, phobos, mass of mars, orbital radius, 9.4x10^6 m, time period, 2.8x10^4 s, kepler’s law, gravitational constant, 6.27x10^23 kg.
