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A man weighs 60 kg at earth's surface. At what height above the earth's surface his weight becomes 30 kg?
(radius of earth = 6400 km).
- a)1624 km
- b)2424 km
- c)2624 km
- d)2826 km
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A man weighs 60 kg at earth's surface. At what height above the ea...
To solve this problem, we can use the concept of gravitational force and the formula for weight.
The weight of an object is given by the formula:
Weight = mass × acceleration due to gravity
At the surface of the Earth, the weight of the man is 60 kg. We can assume the acceleration due to gravity to be approximately 9.8 m/s².
So, at the surface of the Earth:
Weight = 60 kg × 9.8 m/s²
Weight = 588 N
Now, let's consider the height above the Earth's surface where the weight becomes 30 kg. We can assume that the acceleration due to gravity remains constant.
Let's denote the radius of the Earth as R (6400 km or 6400,000 m).
At a distance h above the Earth's surface, the radius becomes R + h.
Using the formula for gravitational force, we have:
Weight = mass × acceleration due to gravity
30 kg × 9.8 m/s² = mass × (gravitational constant × mass of Earth / (R + h)²)
Simplifying the equation, we get:
30 kg × 9.8 m/s² = mass × (9.8 m/s² × (4π² × (R + h)³) / (24 × 60 × 60)²)
Cancelling out the common terms, we have:
1 = (4π² × (R + h)³) / (24 × 60 × 60)²
Rearranging the equation, we get:
(R + h)³ = (24 × 60 × 60)² / (4π²)
Taking the cube root of both sides, we have:
R + h = (24 × 60 × 60)^(2/3) / (4π²)^(1/3)
Substituting the given value of R = 6400 km (or 6400,000 m) and solving for h, we get:
h = [(24 × 60 × 60)^(2/3) / (4π²)^(1/3)] - R
Calculating the value, we find:
h ≈ 2624 km
Therefore, the correct answer is option C) 2624 km.
The weight of an object is given by the formula:
Weight = mass × acceleration due to gravity
At the surface of the Earth, the weight of the man is 60 kg. We can assume the acceleration due to gravity to be approximately 9.8 m/s².
So, at the surface of the Earth:
Weight = 60 kg × 9.8 m/s²
Weight = 588 N
Now, let's consider the height above the Earth's surface where the weight becomes 30 kg. We can assume that the acceleration due to gravity remains constant.
Let's denote the radius of the Earth as R (6400 km or 6400,000 m).
At a distance h above the Earth's surface, the radius becomes R + h.
Using the formula for gravitational force, we have:
Weight = mass × acceleration due to gravity
30 kg × 9.8 m/s² = mass × (gravitational constant × mass of Earth / (R + h)²)
Simplifying the equation, we get:
30 kg × 9.8 m/s² = mass × (9.8 m/s² × (4π² × (R + h)³) / (24 × 60 × 60)²)
Cancelling out the common terms, we have:
1 = (4π² × (R + h)³) / (24 × 60 × 60)²
Rearranging the equation, we get:
(R + h)³ = (24 × 60 × 60)² / (4π²)
Taking the cube root of both sides, we have:
R + h = (24 × 60 × 60)^(2/3) / (4π²)^(1/3)
Substituting the given value of R = 6400 km (or 6400,000 m) and solving for h, we get:
h = [(24 × 60 × 60)^(2/3) / (4π²)^(1/3)] - R
Calculating the value, we find:
h ≈ 2624 km
Therefore, the correct answer is option C) 2624 km.
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Community Answer
A man weighs 60 kg at earth's surface. At what height above the ea...
Actually i am not sure of the answer. The data given in the question is wrong. We shall assume the weight to be 60 kgf at the surface and 30 kgf at height ‘h’. radius of earth is r = 6400 km = 6400000 m. So, weight at surface is = 60 kgf = 60 × 9.8 N Weight at height is = 30 kgf = 30 × 9.8 N If ‘m’ is the mass of the man then on the surface mg = 60 × 9.8 N => m = 60 kg At the height, the weight becomes mg/ = 30 × 9.8 N => g/ = (30 × 9.8)/60 = 4.9 m/s2 We know, the change in ‘g’ with height is given by, g/ = g(1 – 2h/r) => 4.9 = 9.8(1 – 2h/6400000) => h = 1600000 m = 1600 km
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A man weighs 60 kg at earth's surface. At what height above the earth's surface his weight becomes 30 kg?(radius of earth = 6400 km).a)1624 kmb)2424 kmc)2624 kmd)2826 kmCorrect answer is option 'C'. Can you explain this answer?
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