Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Take the mean distance of the moon and the sun from the earth to be

Take the mean distance of the moon and
the sun from the earth to be 0.4\times10 ^{6} km
and 150 \times10 ^{6}  km respectively. Their
masses are 8 \times10 ^{22} kg and 2 \times10 ^{30} kg
respectively. The radius of the earth is
6400 km. Let \Delta F_{1} be the difference in the
forces exerted by the moon at the nearest
and farthest points on the earth and  \Delta F_{2}  be the difference in the force exerted by
the sun at the nearest and farthest points
on the earth. Then, the number closest to \frac{\Delta F_{1}}{\Delta F_{2}}  is :

Answers (1)

@Aravind gupta. watch this

https://www.youtube.com/watch?v=FVOVZOclDtU

\Delta f_{1}= \frac{-2GM_{e}m}{r_{1}^{3}} \Delta r1 \ \ and \ \ \Delta f_{2}=\frac{-2GM_{e}M_{s}}{r_{2}^{3}} \Delta r_{2}

\frac{\Delta f_{1}}{\Delta f_{2}} = \frac{m\Delta r}{r_{1}^{3}}\cdot \frac{r_{2}^{3}}{M_{s}\Delta r_{2}}=\frac{m}{M_{s}}\cdot \frac{r_{2}^{3}}{r_{1}^{3}}\cdot \frac{\Delta r_{1}}{\Delta r_{2}}

using \Delta r_{1} = \Delta r_{2} = 2 R_{earth} \ \ \ m=8\times 10^{22}kg, M_{s}=2\times 10^{30}kg

r_{1} = 0.4\times 10^{6}km, r_{2}=150\times 10^{6}km

we get \frac{\Delta f_{1}}{\Delta f_{2}}=2

Posted by

Safeer PP

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JoSAA Counselling 2025 LIVE: BTech registration starts at josaa.nic.in for IIT, NIT, IIIT admissions
anam-khan

Kota: 17-year old student from UP preparing for JEE dies of electrocution in hostel
anam-khan

CSAB 2025 counselling dates out for special, supernumerary rounds through JEE Mains

Latest Question