A satellite is orbiting the Earth in a circular orbit at a height where the acceleration due to gravity is 4.9 m/s². If the radiu…

Question

A satellite is orbiting the Earth in a circular orbit at a height where the acceleration due to gravity is 4.9 m/s². If the radius of the Earth is 6,371 km, what is the orbital speed of the satellite? (Use g = 9.8 m/s², √2 = 1.414 and √10 = 3.162)

1. A. 4.6 km/s
2. B. 6.6 km/s (Correct answer)
3. C. 9.8 km/s
4. D. 11.2 km/s

Correct Answer

Option B — 6.6 km/s

Detailed Solution & Explanation

The correct answer is 6.6 km/s.

Key Points

• Step 1 — find the orbital radius. Acceleration due to gravity at a distance $r$ from the Earth's centre is $g' = \dfrac{GM}{r^2}$, and at the surface $g = \dfrac{GM}{R^2}$.
  • Given $g' = 4.9 = \dfrac{g}{2}$:
  • $\frac{GM}{r^2} = \frac{1}{2}\cdot\frac{GM}{R^2} \;\Rightarrow\; r^2 = 2R^2 \;\Rightarrow\; r = R\sqrt{2}$
  • $r = 6371 \times 1.414 = 9008.6 \text{ km} = 9.0086 \times 10^6 \text{ m}$
• Step 2 — apply the orbital speed condition. For a circular orbit, gravity supplies the centripetal force:
  • $\frac{mv^2}{r} = mg' \;\Rightarrow\; v = \sqrt{g' r}$
  • $v = \sqrt{4.9 \times 9.0086 \times 10^6} = \sqrt{4.414 \times 10^7} \approx 6644 \text{ m/s}$
• $v \approx \mathbf{6.6 \text{ km/s}}$

Shortcut Trick

• Since $g' = \frac{g}{2}$ implies $r = R\sqrt{2}$, the orbital speed reduces to:
  • $v = \sqrt{\frac{g}{2} \times R\sqrt{2}} = \sqrt{\frac{gR}{\sqrt{2}}}$
• With $gR = 9.8 \times 6.371 \times 10^6 = 6.24 \times 10^7$, dividing by $1.414$ gives $4.41 \times 10^7$, whose square root is about 6.6 km/s.

Additional Information

• The three cosmic speeds worth remembering:
  • Orbital speed at the Earth's surface $v_o = \sqrt{gR} \approx 7.9$ km/s — the first cosmic speed.
  • Escape velocity $v_e = \sqrt{2gR} \approx 11.2$ km/s $= \sqrt{2}\,v_o$.
  • Orbital speed decreases as altitude increases, which is why 6.6 km/s is below the surface value of 7.9 km/s — a useful sanity check that immediately rejects the 9.8 and 11.2 options.
• Note that 9.8 km/s is a distractor built from the numerical value of $g$, and 11.2 km/s is the escape velocity — neither is an orbital speed.