Question

The distance between centre of the earth and moon is 384000 km. If the mass of the earth is 6×10²⁴ kg and G=6.67×10⁻¹¹ Nm² / kg². The speed of the moon is nearly?

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\sf\implies The \ Distance\ between \ centres \ of \ earth \ and \ moon \ is \ 384000km .\\\sf\implies The \ mass \ of \ the \ earth \ is \ 3 \times 10^{24} kg .\\\sf\implies The \ value \ of \ G \ is \ 6.67 \times 10^{-11} Nm^2/ kg^2.$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\sf\implies The \ orbital \ speed \ of \ moon .$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

The speed needed to attain the balance between inertia of the satellite's motion and the gravity's pull on the satellite is called Orbital Speed .Orbital speed is given by :-

$\qquad\purple{\bigstar}\boxed{\boxed{\green{\bf v_{orbital}= \sqrt{\dfrac{GM_{earth}}{R}}}}}\purple{\bigstar}$

$\underline{\pink{\boldsymbol{Using \ this \ formula \ we \ have :- }}}$

$\sf:\implies v_{orbital}=\sqrt{\dfrac{GM_{earth}}{R}}\\\\\sf:\implies v_{orbital} = \sqrt{\dfrac{6.67\times 10^{-11} Nm^2/kg^2 \times 6 \times 10^{24} kg }{384000 km}}\\\\\sf:\implies v_{orbital} = \sqrt{\dfrac{6.67\times 10^{-11} Nm^2/kg^2 \times 6 \times 10^{24} kg }{384\times 10^6 m}} \\\\\sf:\implies v_{orbital} = \sqrt{\dfrac{39.967\times 10^{18}\times 10^{-11} }{384}} \\\\\sf:\implies v_{orbital} = \sqrt{0.104\times 10^{7} m/s } \\\\\sf:\implies v_{orbital } = \sqrt{ 1.04 \times 10^6 m/s }\\\\\sf:\implies v_{orbital} = 1 \times 1000 m /s \\\\\sf:\implies \boxed{\orange{\sf v_{orbital} = 1km/s (approx.)}}$

$\underline{\blue{\sf Hence \ the \ orbital \ speed \ is\:\: \textsf{\textbf{\blue{1km/s }}}}}$.