Gravitation: PYQ Walkthrough (8)

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Tags Gravitation

Question

(JEE Main pattern, recurring 2022–2024) A satellite is orbiting Earth at a height equal to the radius of the Earth RR. Find its orbital velocity and time period. Take g=9.8g = 9.8 m/s2^2 at Earth’s surface and R=6.4×106R = 6.4 \times 10^6 m.

Solution — Step by Step

The satellite is at height h=Rh = R above the surface. So the orbital radius (measured from Earth’s centre) is r=R+h=2Rr = R + h = 2R. This is the distance we plug into the gravitational equations — not hh alone.

For a circular orbit, gravity provides centripetal force:

GMmr2=mv2r  ⟹  v=GMr\frac{GMm}{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac{GM}{r}}

Replace GMGM using g=GM/R2g = GM/R^2, so GM=gR2GM = gR^2.

v=gR22R=gR2v = \sqrt{\frac{gR^2}{2R}} = \sqrt{\frac{gR}{2}} v=9.8×6.4×1062=3.136×107≈5.6×103 m/sv = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{2}} = \sqrt{3.136 \times 10^7} \approx 5.6 \times 10^3 \text{ m/s}

So v≈5.6v \approx 5.6 km/s.

 T=2πrv=2π×2Rv=4πRvT = \frac{2\pi r}{v} = \frac{2\pi \times 2R}{v} = \frac{4\pi R}{v} T=4π×6.4×1065.6×103≈1.43×104 s≈4 hoursT = \frac{4\pi \times 6.4 \times 10^6}{5.6 \times 10^3} \approx 1.43 \times 10^4 \text{ s} \approx 4 \text{ hours}

Final Answer: v≈5.6v \approx 5.6 km/s, T≈4T \approx 4 hours.

Why This Works

Replacing GMGM with gR2gR^2 is the single most useful trick in gravitation problems. JEE and NEET rarely give you GG and the mass of Earth directly — they give you gg at the surface, and you must do this substitution.

The orbital velocity decreases as rr increases, which is why higher satellites move slower (longer period). Geostationary satellites at r≈6.6Rr \approx 6.6R have T=24T = 24 hours.

Alternative Method

Use T2∝r3T^2 \propto r^3 (Kepler’s third law). For a low-Earth orbit at r=Rr = R, T0=2πR/g≈84T_0 = 2\pi\sqrt{R/g} \approx 84 minutes. Scale up: T=T0⋅(r/R)3/2=84⋅21.5≈237T = T_0 \cdot (r/R)^{3/2} = 84 \cdot 2^{1.5} \approx 237 min ≈4\approx 4 hours. Faster than computing from scratch.

The most common slip: using r=h=Rr = h = R instead of r=R+h=2Rr = R + h = 2R. The rr in v=GM/rv = \sqrt{GM/r} is the distance from Earth’s centre, not from its surface. This single error gives a wrong orbital velocity by a factor of 2\sqrt{2}.

For PYQs, “satellite at height RR” or “satellite at twice the Earth’s radius” both mean r=2Rr = 2R. Read carefully. Examiners flip between height-based and radius-based phrasing every year.

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