Physics Q93 – AP-EAMCET 2026

If the gravitational force acting on a body due to earth at a height equal to twice the radius of the earth is

30 N

, then the mass of the body is (Acceleration due to gravity on the surface of the earth =

10 {ms}^{- 2}

)

Gravitational force on a body is given by the product of its mass and the acceleration due to gravity at that specific location.

Step 1: Determine the acceleration due to gravity at the given height.✦ Active

The acceleration due to gravity at a height $h$ above the Earth's surface is given by:

g_{h} = g_{s} {(\frac{R_{e}}{R_{e} + h})}^{2}

Given that the height $h$ is twice the radius of the Earth, $h = 2 R_{e}$ . Substituting this into the formula:

g_{h} = g_{s} {(\frac{R_{e}}{R_{e} + 2 R_{e}})}^{2} = g_{s} {(\frac{R_{e}}{3 R_{e}})}^{2} = g_{s} {(\frac{1}{3})}^{2} = \frac{g_{s}}{9}

We are given $g_{s} = 10 {ms}^{- 2}$ , so:

g_{h} = \frac{10}{9} {ms}^{- 2}

💡 Teacher's Secret Hint

Remember that the acceleration due to gravity decreases with height, following an inverse square law for distance from the center of the Earth.

Step 2: Calculate the mass of the body.○ Expand

The gravitational force $F_{h}$ acting on the body at this height is given by:

F_{h} = m g_{h}

We are given $F_{h} = 30 N$ and we calculated $g_{h} = \frac{10}{9} {ms}^{- 2}$ . We can now solve for the mass $m$ :

30 N = m \times \frac{10}{9} {ms}^{- 2}

m = \frac{30 \times 9}{10} kg

m = 3 \times 9 kg

m = 27 kg

💡 Teacher's Secret Hint

Ensure consistent units throughout the calculation. Here, force is in Newtons, acceleration in ${ms}^{- 2}$ , so mass will be in kilograms.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 93 – AP-EAMCET 2026