Absolutely! Let's delve into this interesting physics problem together.
Question:
The mass of planet B is twice the mass of planet A, but its radius is half that of planet A. Similarly, the mass of planet C is half the mass of planet A, but its radius is twice that of planet A. If the weight of an object on planets A, B, and C are W₁, W₂, and W₃ respectively, which of the following orders is correct?
Options:
i. W₁ > W₃ > W₂
ii. W₂ > W₁ > W₃
iii. W₁ > W₂ > W₃
iv. W₂ > W₃ > W₁
Answer:
ii. W₂ > W₁ > W₃
Explanation:
To determine the correct order of weights, we'll compare the gravitational acceleration on each planet since weight depends on gravity.
Understanding Weight and Gravity:
-
Weight (W) of an object on a planet is given by:
[
W = m \times g
]- ( m ) = mass of the object (constant for all planets)
- ( g ) = acceleration due to gravity on the planet
-
Acceleration due to gravity (g) is calculated by:
[
g = \frac{G \times M}{R^2}
]- ( G ) = gravitational constant
- ( M ) = mass of the planet
- ( R ) = radius of the planet
Calculations:
Let's denote:
-
Planet A:
- Mass ( M_A = M )
- Radius ( R_A = R )
- Gravity ( g_A = \frac{G \times M}{R^2} )
-
Planet B:
- Mass ( M_B = 2M ) (twice of A)
- Radius ( R_B = \frac{R}{2} ) (half of A)
- Gravity ( g_B = \frac{G \times M_B}{R_B^2} )
[
g_B = \frac{G \times (2M)}{\left( \frac{R}{2} \right)^2} = \frac{G \times 2M}{\frac{R^2}{4}} = \frac{8G \times M}{R^2} = 8g_A
]
-
Planet C:
- Mass ( M_C = \frac{M}{2} ) (half of A)
- Radius ( R_C = 2R ) (twice of A)
- Gravity ( g_C = \frac{G \times M_C}{R_C^2} )
[
g_C = \frac{G \times \left( \frac{M}{2} \right)}{(2R)^2} = \frac{G \times \frac{M}{2}}{4R^2} = \frac{G \times M}{8R^2} = \frac{g_A}{8}
]
Comparing Weights:
1
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