A planet's moon travels in an approximately circular orbit of radius 7.0 ✕ 10⁷ m with a period of 6 h 38 min. Calculate the mass of the planet from this information. ___ kg

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Answer:

3.56×10²⁶ Kg.

Explanation:

Note: The gravitational force is acting as the centripetal force.

Fg = Fc........................... Equation 1

Where Fg = gravitational Force, Fc = centripetal force.

Recall,

Fg = GMm/r²......................... Equation 2

Fc = mv²/r............................. Equation 3

Where M = mass of the planet, m = mass of the moon, r = radius of the orbit and G = Universal gravitational constant.

Substituting equation 2 and 3 into equation 1

GMm/r² = mv²/r

Simplifying the equation above,

M = v²r/G .............................. Equation 4.

The period of the moon in the orbit

T = 2πr/v

Making v the subject of the equation,

v = 2πr/T............................. Equation 5

where r = 7.0×10⁷ m, T = 6 h 38 min = (6×3600 + 38×60) s = (21600+2280) s

T = 23880 s, π = 3.14

v = (2×3.14×7.0×10⁷ )/23880

v = 18409 m/s

Also Given: G = 6.67×10⁻¹¹ Nm²/kg²

Also substituting into equation 4

M = 18409²×7.0×10⁷ /(6.67×10⁻¹¹)

M = 3.56×10²⁶ Kg.

Thus the mass of the planet =  3.56×10²⁶ Kg.

The mass of the planet is required.

The mass of the planet is [tex]3.56\times 10^{26}\ \text{kg}[/tex]

M = Mass of planet

r = Radius of orbit = [tex]7\times 10^7\ \text{m}[/tex]

t = Time period = [tex]6\times 60\times 60+38\times 60=23880\ \text{s}[/tex]

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

Mass is given by

[tex]M=\dfrac{4\pi^2r^3}{t^2G}\\\Rightarrow M=\dfrac{4\pi^2\times (7\times 10^7)^3}{23880^2\times 6.674\times 10^{-11}}\\\Rightarrow M=3.56\times 10^{26}\ \text{kg}[/tex]

The mass of the planet is [tex]3.56\times 10^{26}\ \text{kg}[/tex]

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