A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f1V (f1 < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f2V. Then:
f2 = f1
Understanding Buoyancy and Submerged Volume
This problem asks us to compare the fraction of a sphere's volume submerged when floating in water under two different conditions: on Earth under gravity and on an accelerating spaceship in outer space.
The key principle here is that for an object to float, the buoyant force acting on it must be equal in magnitude to its weight. The buoyant force is the weight of the fluid displaced by the submerged part of the object.
Let V be the total volume of the sphere, $\rho_s$ be the density of the sphere, and $\rho_w$ be the density of water. The sphere floats because its density is less than the density of water ($\rho_s < \rho_w$).
Case 1: Sphere Floating on Earth
On Earth, the acceleration due to gravity is g.
- Weight of the sphere (WEarth) = Mass × Acceleration = $\rho_s V g$
- Volume submerged = f1V
- Volume of displaced water = f1V
- Weight of displaced water (Buoyant Force, BEarth) = Density of water × Volume of displaced water × Acceleration = $\rho_w (f_1 V) g$
Since the sphere is floating, the buoyant force equals its weight:
BEarth = WEarth
$\rho_w (f_1 V) g = \rho_s V g$
We can cancel out the volume V (since V > 0) and the acceleration g (since g > 0) from both sides:
$\rho_w f_1 = \rho_s$
Thus, the fraction submerged on Earth is:
$f_1 = \frac{\rho_s}{\rho_w}$
Case 2: Sphere Floating on an Accelerating Spaceship
On the spaceship in outer space, there is an acceleration a ($a < g$). This acceleration acts as the effective gravitational acceleration within the spaceship's frame of reference, causing the water to settle and objects to experience weight and buoyancy relative to this acceleration.
- Effective acceleration = a
- Weight of the sphere (WSpaceship) = Mass × Effective acceleration = $\rho_s V a$
- Volume submerged = f2V
- Volume of displaced water = f2V
- Weight of displaced water (Buoyant Force, BSpaceship) = Density of water × Volume of displaced water × Effective acceleration = $\rho_w (f_2 V) a$
Since the sphere is floating on the spaceship, the buoyant force equals its weight:
BSpaceship = WSpaceship
$\rho_w (f_2 V) a = \rho_s V a$
We can cancel out the volume V (since V > 0) and the effective acceleration a (since a > 0) from both sides:
$\rho_w f_2 = \rho_s$
Thus, the fraction submerged on the spaceship is:
$f_2 = \frac{\rho_s}{\rho_w}$
Comparing the Submerged Fractions
From Case 1, we found $f_1 = \frac{\rho_s}{\rho_w}$.
From Case 2, we found $f_2 = \frac{\rho_s}{\rho_w}$.
Therefore, $f_1 = f_2$.
The fraction of the volume submerged when an object floats in a fluid depends only on the ratio of the object's density to the fluid's density. It does not depend on the magnitude of the uniform acceleration (gravity or effective acceleration) acting on the system, as long as that acceleration is non-zero and uniform throughout the fluid and the object.
Conclusion
The submerged volume on the spaceship ($f_2V$) is the same fraction of the total volume as it is on Earth ($f_1V$). Thus, $f_2 = f_1$.
| Condition | Effective Acceleration | Weight of Sphere | Buoyant Force | Floating Condition | Submerged Fraction |
|---|---|---|---|---|---|
| Earth | g | $\rho_s V g$ | $\rho_w (f_1 V) g$ | $\rho_w f_1 g = \rho_s g$ | $f_1 = \frac{\rho_s}{\rho_w}$ |
| Spaceship | a | $\rho_s V a$ | $\rho_w (f_2 V) a$ | $\rho_w f_2 a = \rho_s a$ | $f_2 = \frac{\rho_s}{\rho_w}$ |
The value of acceleration (g or a) cancels out from the equation when determining the fraction of volume submerged.
Revision Table: Buoyancy Concepts
| Concept | Description | Relevant Principle |
|---|---|---|
| Buoyant Force | Upward force exerted by a fluid that opposes the weight of an immersed object. | Archimedes' Principle |
| Archimedes' Principle | The buoyant force on an object equals the weight of the fluid displaced by the object. | $B = \rho_{fluid} V_{submerged} g_{effective}$ |
| Floating | Occurs when the buoyant force equals the object's weight ($B = W$). The object is less dense than the fluid. | $V_{submerged} / V_{total} = \rho_{object} / \rho_{fluid}$ |
| Sinking | Occurs when the buoyant force is less than the object's weight ($B < W$). The object is denser than the fluid. | Object accelerates downwards |
Additional Information: Buoyancy in Different Frames
Understanding buoyancy in different reference frames, like an accelerating spaceship, is a good way to solidify the concepts of inertial frames and effective forces.
- In the accelerating spaceship's frame, the acceleration 'a' acts like a uniform gravitational field. The physics principles, including Archimedes' principle, apply in this non-inertial frame if we include this effective acceleration in calculations of weight and buoyant force.
- Both the weight of the sphere and the weight of the displaced water are directly proportional to the effective acceleration. When setting these two forces equal for the floating condition, the acceleration term cancels out, meaning the submerged fraction is independent of the acceleration's magnitude (as long as it's uniform and non-zero).
- This is analogous to how your weight and the buoyant force on you in an elevator accelerating upwards or downwards both change proportionally, but whether you float or sink in a fluid inside the elevator would still depend on your density relative to the fluid's density, not the elevator's acceleration.
- The density ratio ($\rho_s / \rho_w$) is an intrinsic property of the sphere and the water, independent of the external acceleration field.
Important Questions from Miscellaneous
A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s2 and neglecting air resistance will give :
A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
Consider the following statements:
1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :
A mass is attached to a spring that hangs vertically. The extension produced in the spring is 6 cm on Earth. The acceleration due to gravity on the surface of the Moon is one-sixth of its value on the surface of the Earth. The extension of the spring on the Moon would be:
