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Sagot :
The period of the new planet is 1.053 × 10¹⁰ s
Gravitational force on planet
The gravitational force on the planet, F equals the centripetal force on the planet, F'
F = F'
GMm/R² = mRω² where
- G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²,
- M = mass of sun = 1.989 × 10³⁰ kg,
- m = mass of planet,
- R = radius of planet = 48.1 au = 48.1 × 1.496 × 10¹¹ m = 7.1958 × 10¹² m,
- ω = angular speed of planet = 2π/T where
- T = period of planet
So, GMm/R² = mRω²
GMm/R² = mR(2π/T)²
GM/R³ = (2π)²/T²
Period of the planet
Making T subject of the formula, we have
T = 2π√(R³/GM)
Substituting the values of the variables into the equation, we have
T = 2π√(R³/GM)
T = 2π√((7.1958 × 10¹² m)³/[6.67 × 10⁻¹¹ Nm²/kg² × 1.989 × 10³⁰ kg])
T = 2π√(372.5952 × 10³⁶ m³/13.26663 × 10¹⁹ Nm²/kg)
T = 2π√(28.085 × 10¹⁷ Nm/kg)
T = 2π√(2.8085 × 10¹⁸ Nm/kg)
T = 2π × 1.676 × 10⁹ s
T = 10.53 × 10⁹ s
T = 1.053 × 10¹⁰ s
The period of the new planet is 1.053 × 10¹⁰ s
Learn more about period of planet here:
https://brainly.com/question/15402804
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