JEE Main 25 Feb 2021 Shift 1 Solved Paper

Given,
MA=200kg,MB=400kg,HA=600km,HB=1600km and ‌‌RA=RE+HA=6400+600=7000km RB=RE+HB=6400+1600=8000km
Let TA,TB,ωA,ωB,RA and RB be the time period, angular frequencies and radii of satellites A and B, respectively. Force on satellite A,FA=mAωA2RA=‌

GMmA

RA2

⇒‌‌ωA2=‌

GM

RA3

but, ‌‌ωA=‌

2π

TA

∴‌‌(‌

2π

TA

)2=‌

GM

RA3

⇒TA=√‌

4π2RA3

GM

Similarly, ‌‌TB=√‌

4π2RB3

GM

∴‌‌TB−TA‌‌=√‌

4π2

GM

(√RB3−√RA3)
‌‌=‌

2π

√GM

[√(8×103)3−√(7×103)3]
=‌

2π×109

√GM

(8√8−7√7)
=‌

2π×109

√6.67×10−11×6×1024

(4.107)
=‌

2π

√4×1014

×109(4.107)
=π×102×4.107=12.9×102s
=1.33×103s