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A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth ?
- a)30 N
- b)24 N
- c)48 N
- d)32 N
Correct answer is option 'D'. Can you explain this answer?
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A body weighs 72 N on the surface of the earth. What is the gravitatio...
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A body weighs 72 N on the surface of the earth. What is the gravitatio...
Given: Weight of body on surface of Earth = 72 N
Height above surface of Earth = r/2 (where r is the radius of Earth)
To find: Gravitational force on the body at this height
Formula: Gravitational force = (GMm)/r^2
where G is the gravitational constant, M is the mass of Earth, m is the mass of the body, and r is the distance between the center of Earth and the body
Approach:
1. Find the radius of Earth (r) using the formula for volume of a sphere: (4/3)πr^3 = V, where V is the volume of Earth
2. Find the distance between the body and the center of Earth (r') using the Pythagorean theorem: r' = r + r/2 = 3r/2
3. Find the mass of Earth (M) using the formula for density: ρ = M/V, where ρ is the density of Earth
4. Substitute the values in the formula for gravitational force and simplify
Calculation:
1. Radius of Earth, r = [(3V)/(4π)]^(1/3) = [(3*4/3*π*r^3)/(4π)]^(1/3) = r
2. Distance between body and center of Earth, r' = r + r/2 = 3r/2
3. Mass of Earth, M = ρV = ρ(4/3)πr^3 = (5.5 g/cm^3)(4/3)π(6400 km)^3 = 5.97 x 10^24 kg
4. Gravitational force, F = (GMm)/r'^2 = (6.67 x 10^-11 Nm^2/kg^2)(5.97 x 10^24 kg)(m)/(3r/2)^2
Given weight of body on surface of Earth = 72 N
Weight of body = mg, where g is acceleration due to gravity on surface of Earth (9.8 m/s^2)
Therefore, mass of body, m = 72 N/9.8 m/s^2 = 7.35 kg
Substituting the values:
F = (6.67 x 10^-11 Nm^2/kg^2)(5.97 x 10^24 kg)(7.35 kg)/(3r/2)^2
F = 31.77 N ≈ 32 N
Therefore, the gravitational force on the body at a height equal to half the radius of Earth is approximately 32 N.
Answer: d)32 N
Height above surface of Earth = r/2 (where r is the radius of Earth)
To find: Gravitational force on the body at this height
Formula: Gravitational force = (GMm)/r^2
where G is the gravitational constant, M is the mass of Earth, m is the mass of the body, and r is the distance between the center of Earth and the body
Approach:
1. Find the radius of Earth (r) using the formula for volume of a sphere: (4/3)πr^3 = V, where V is the volume of Earth
2. Find the distance between the body and the center of Earth (r') using the Pythagorean theorem: r' = r + r/2 = 3r/2
3. Find the mass of Earth (M) using the formula for density: ρ = M/V, where ρ is the density of Earth
4. Substitute the values in the formula for gravitational force and simplify
Calculation:
1. Radius of Earth, r = [(3V)/(4π)]^(1/3) = [(3*4/3*π*r^3)/(4π)]^(1/3) = r
2. Distance between body and center of Earth, r' = r + r/2 = 3r/2
3. Mass of Earth, M = ρV = ρ(4/3)πr^3 = (5.5 g/cm^3)(4/3)π(6400 km)^3 = 5.97 x 10^24 kg
4. Gravitational force, F = (GMm)/r'^2 = (6.67 x 10^-11 Nm^2/kg^2)(5.97 x 10^24 kg)(m)/(3r/2)^2
Given weight of body on surface of Earth = 72 N
Weight of body = mg, where g is acceleration due to gravity on surface of Earth (9.8 m/s^2)
Therefore, mass of body, m = 72 N/9.8 m/s^2 = 7.35 kg
Substituting the values:
F = (6.67 x 10^-11 Nm^2/kg^2)(5.97 x 10^24 kg)(7.35 kg)/(3r/2)^2
F = 31.77 N ≈ 32 N
Therefore, the gravitational force on the body at a height equal to half the radius of Earth is approximately 32 N.
Answer: d)32 N
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A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth ?a)30 Nb)24 Nc)48 Nd)32 NCorrect answer is option 'D'. Can you explain this answer?
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