
The moon is about $ 3.8 \times {10^5}Km $ from the center of the earth. It takes about $ 27days $ for complete the orbit around the earth. Calculate the speed of the moon in Km per day.
Answer
563.1k+ views
Hint: Here in this question we have to find the speed of the moon in Km per day and for this, we will use the formula given by time, $ T = \dfrac{{2\pi R}}{V} $ and by substituting the values and solving it we will get the velocity for per day.
Formula used
Time to complete an orbit is given by
$ T = \dfrac{{2\pi R}}{V} $
Here, $ T $ , will be the time,
$ R $ , will be the radius of the moon from the earth,
$ V $ , will be the velocity.
Complete step by step answer:
So in this question, we have the radius given by $ 3.8 \times {10^5}Km $ and the time is given as $ 27days $ . So now for calculating the time, by using the formula we will get
$ \Rightarrow T = \dfrac{{2\pi R}}{V} $
Now on substituting the values, we will get the equation as
$ \Rightarrow 27days = \dfrac{{2\pi \times \left( {3.8 \times {{10}^5}} \right)km}}{V} $
And from here the velocity will be given by
$ \Rightarrow V = \dfrac{{2\pi \times \left( {3.8 \times {{10}^5}} \right)km}}{{27days}} $
As we the value of $ \pi = 3.14 $ , so on substituting the values we will get the equation as
$ \Rightarrow V = \dfrac{{2 \times 3.14\left( {3.8 \times {{10}^5}} \right)km}}{{27days}} $
Now on solving the multiplication and division, we will get the equation as
$ \Rightarrow V = 8.4 \times {10^4}km/days $
Therefore, the speed on the moon will be equal to $ 8.4 \times {10^4}km/days $ .
Note:
In this type of question it becomes important to memorize the formula and also by the statement of the question we have to check whether we have the diameter given or the radius given and according to the use we can take it from there. Sometimes the value of pie will be given in the question, and if it is given then we have to take that value only otherwise any one of the two.
Formula used
Time to complete an orbit is given by
$ T = \dfrac{{2\pi R}}{V} $
Here, $ T $ , will be the time,
$ R $ , will be the radius of the moon from the earth,
$ V $ , will be the velocity.
Complete step by step answer:
So in this question, we have the radius given by $ 3.8 \times {10^5}Km $ and the time is given as $ 27days $ . So now for calculating the time, by using the formula we will get
$ \Rightarrow T = \dfrac{{2\pi R}}{V} $
Now on substituting the values, we will get the equation as
$ \Rightarrow 27days = \dfrac{{2\pi \times \left( {3.8 \times {{10}^5}} \right)km}}{V} $
And from here the velocity will be given by
$ \Rightarrow V = \dfrac{{2\pi \times \left( {3.8 \times {{10}^5}} \right)km}}{{27days}} $
As we the value of $ \pi = 3.14 $ , so on substituting the values we will get the equation as
$ \Rightarrow V = \dfrac{{2 \times 3.14\left( {3.8 \times {{10}^5}} \right)km}}{{27days}} $
Now on solving the multiplication and division, we will get the equation as
$ \Rightarrow V = 8.4 \times {10^4}km/days $
Therefore, the speed on the moon will be equal to $ 8.4 \times {10^4}km/days $ .
Note:
In this type of question it becomes important to memorize the formula and also by the statement of the question we have to check whether we have the diameter given or the radius given and according to the use we can take it from there. Sometimes the value of pie will be given in the question, and if it is given then we have to take that value only otherwise any one of the two.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

Why is 1 molar aqueous solution more concentrated than class 11 chemistry CBSE

SiO2GeO2 SnOand PbOare respectively A acidic amphoteric class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

