A planet has double the mass of the earth. Its average density is equal to
that of the earth. An object weighing W on earth will weigh on that planet:
A planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing W on earth will weigh on that planet:
1
(2)\(^{\frac{1}{3}}\)
(2)\(^{-\frac{1}{3}}\)
2
The Correct Option is B
Approach Solution - 1
The correct answer is option (B): \((2) ^{\frac{1}{3}}\)
Approach Solution -2
Weight on a Different Planet Problem
Step 1: Relationship between Mass, Density, and Radius
Let \( M_e, R_e, \) and \( \rho_e \) be the mass, radius, and average density of Earth, respectively. Let \( M_p, R_p, \) and \( \rho_p \) be the mass, radius, and average density of the planet, respectively. Given that \( M_p = 2M_e \) and \( \rho_p = \rho_e \). Density is defined as mass divided by volume: \( \rho = \frac{M}{V} \). Assuming spherical shapes, \( V = \frac{4}{3} \pi R^3 \). Thus:
\( \rho_e = \frac{M_e}{\frac{4}{3}\pi R_e^3} \) and \( \rho_p = \frac{M_p}{\frac{4}{3}\pi R_p^3} \)
Since \( \rho_e = \rho_p \):
\( \frac{M_e}{R_e^3} = \frac{M_p}{R_p^3} \)
\( \frac{M_e}{R_e^3} = \frac{2M_e}{R_p^3} \)
\( R_p^3 = 2R_e^3 \implies R_p = 2^{1/3} R_e \)
Step 2: Relationship between Weight and Gravitational Acceleration
Weight (W) is given by \( W = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.
Step 3: Gravitational Acceleration
The acceleration due to gravity (g) is given by:
\( g = \frac{GM}{R^2} \)
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius. Let \( g_e \) and \( g_p \) be the acceleration due to gravity on Earth and the planet, respectively.
\( \frac{g_e}{g_p} = \frac{M_e / R_e^2}{M_p / R_p^2} = \frac{M_e}{R_e^2} \times \frac{R_p^2}{M_p} = \frac{M_e}{R_e^2} \times \frac{(2^{1/3} R_e)^2}{2M_e} = \frac{2^{2/3}}{2} = 2^{-1/3} \)
\( g_p = 2^{1/3} g_e \)
Step 4: Weight on the Planet
Since the mass of the object remains constant, the weight on the planet (\( W_p \)) is:
\( W_p = mg_p = m(2^{1/3} g_e) = 2^{1/3} (mg_e) = 2^{1/3} W \)
Conclusion:
The object will weigh \( \mathbf{2^{1/3}W} \) on the planet (Option 1).
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Concepts Used:
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- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].