Question:

A research satellite of mass $200\mathrm{kg}$ circles the earth in an orbit of average radius $\frac{3R}{2}$, where $R$ is the radius of the earth. Assuming the gravitational pull on a mass of $1\mathrm{kg}$ on the earth's surface to be $10\mathrm{N}$, the pull on the satellite will be

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Gravitational force varies as $1/r^2$. At height $h$, $g' = \frac{gR^2}{(R+h)^2}$.
Updated On: Apr 8, 2026
  • $880\mathrm{N}$
  • $889\mathrm{N}$
  • $890\mathrm{N}$
  • $892\mathrm{N}$
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Gravitational pull (weight) at earth's surface for 1 kg = $g = 10$ N/kg. So $GM = gR^2 = 10R^2$.
Step 2: Detailed Explanation:
Pull on satellite = $\frac{GMm}{r^2} = \frac{10R^2 \times 200}{(3R/2)^2} = \frac{2000R^2}{9R^2/4} = \frac{2000 \times 4}{9} = \frac{8000}{9} \approx 888.9$ N.
Step 3: Final Answer:
The pull on the satellite is approximately $889$ N.
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