QJEEM25Phdee2: JEE Main 2025 Physics - Class 11 JEEM25Phdee2 Solution: A satellite is launched into a circular orbit of radius ' R '…

$T \propto R^{\frac{3}{2}}$

For a satellite in a circular orbit with radius $R$ , the time period is given by

$T = 2\pi \sqrt{\frac{R^3}{GM}}.$

If the radius is increased to $1.03R$ , the new time period $T'$ becomes

$T' = 2\pi \sqrt{\frac{(1.03R)^3}{GM}} = T \times (1.03)^{\frac{3}{2}}.$

For a small change, we can approximate

$(1.03)^{\frac{3}{2}} \approx 1 + \frac{3}{2} \times 0.03 = 1 + 0.045 = 1.045.$

This means the new time period is approximately 4.5% larger than the original time period.

Thus, the time period of the second satellite is larger by approximately 4.5%.