Suppose a small planet is discovered that is 14 times far from the sun as the Earth- The distance of earth from the sun 1-5 - 1011 m - Use Kepler-s law of harmonies to predict the orbital period of such a planet- GIVEN- T 2- R 3-2-97 - 10-19 s 2 - m 3A- 52-4 yearsB- 60 yearsC- 79-8 yearsD- 36-2 year

The correct option is A 52.4 yearsUse Kepler's third law: T_e^2/R_e^3=T_p^2/R_p^3 Rearranging to solve for T_p: (T_p)^2= (T_e^2/R_e^3 )× (R_p)^3 or (T_p)^2=(T ...

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Standard IX

Physics

Earth, Moon and High Tide

Suppose a sma...

Question

Suppose a small planet is discovered that is 14 times far from the sun as the Earth. The distance of earth from the sun $(1.5 \times 10^{11} m)$ . Use Kepler's law of harmonies to predict the orbital period of such a planet. GIVEN: $\frac{T^{2}}{R^{3}} = 2.97 \times 10^{- 19} s^{2} / m^{3}$

52.4 years

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60 years

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79.8 years

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36.2 year

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Solution

The correct option is A 52.4 years
Use Kepler's third law:
$\frac{T_{e}^{2}}{R_{e}^{3}} = \frac{T_{p}^{2}}{R_{p}^{3}}$
Rearranging to solve for $T_{p}$ :
$(T_{p})^{2} = (\frac{T_{e}^{2}}{R_{e}^{3}}) \times (R_{p})^{3}$
or $(T_{p})^{2} = (T_{e})^{2} \times {(\frac{R_{p}}{R_{e}})}^{3} w h e r e \frac{R_{p}}{R_{e}} = 14$
so $(T_{p})^{2} = (T_{e})^{2} \times (14)^{3} w h e r e T_{e} = 1 y r$
$(T_{p})^{2} = (1 y r)^{2} \times (14)^{3} = 2744 y r^{2}$
$T_{p} = \sqrt{(2744 y r^{2})}$
$T_{p l a n e t} = 52.4 y e a r$

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Suppose a small planet is discovered that is 14 times far from the sun as the Earth. The distance of earth from the sun (1.5×1011m). Use Kepler's law of harmonies to predict the orbital period of such a planet. GIVEN: T2R3=2.97×10−19s2/m3

The correct option is A 52.4 yearsUse Kepler's third law: T2eR3e=T2pR3p Rearranging to solve for Tp: (Tp)2=(T2eR3e)×(Rp)3 or (Tp)2=(Te)2×(RpRe)3 where RpRe=14 so (Tp)2=(Te)2×(14)3 where Te=1 yr(Tp)2=(1 yr)2×(14)3=2744 yr2Tp=√(2744 yr2)Tplanet=52.4 year

Suppose a small planet is discovered that is 14 times far from the sun as the Earth. The distance of earth from the sun $(1.5 \times 10^{11} m)$ . Use Kepler's law of harmonies to predict the orbital period of such a planet. GIVEN: $\frac{T^{2}}{R^{3}} = 2.97 \times 10^{- 19} s^{2} / m^{3}$