MCQ on Gravitation

Explanation: The question asks how escape velocity changes when both the Mass and radius of a planet are doubled compared to Earth. Escape velocity depends on the gravitational pull a planet exerts on an object trying to leave its surface. It is determined by the relation v ∝ √(M/R), where M is Mass and R is radius.

When both Mass and radius are scaled by the same factor, their ratio becomes important. In this case, Mass becomes 2M and radius becomes 2R. Substituting into the proportional expression, we get √(2M / 2R). The factor of 2 cancels out, leaving the same ratio as before.

This means the escape velocity remains unchanged despite the increase in both parameters. The gravitational field strength at the surface effectively stays proportional in such a way that the required escape speed does not change.

Think of it like stretching both the gravitational pull and the distance equally—the balance between them stays the same.

In summary, when both Mass and radius are increased proportionally, escape velocity remains unchanged because it depends on their ratio, not their absolute values.

Explanation: The question involves finding how gravitational potential energy changes when a body is moved from Earth’s surface to a height equal to one-fifth of Earth’s radius. Unlike small height changes, this requires using the concept of gravitational potential energy in a varying gravitational field.

Gravitational potential energy near Earth is commonly approximated as mgh, but this only works when height is very small compared to Earth’s radius. For larger heights, the correct expression depends on the inverse relationship with distance from Earth’s center.

Initially, the object is at a distance R from Earth’s center, and after rising, it is at a distance R + R/5 = 6R/5. The potential energy difference depends on the change in inverse distance terms. By comparing these values, one can determine how much energy is gained.

This situation is similar to lifting an object where gravity weakens with height, so the energy gained is less than what a simple mgh estimate would suggest.

In summary, when height is comparable to Earth’s radius, gravitational potential energy must be calculated using distance-based expressions, leading to a reduced increase compared to constant-g assumptions.

Explanation: This question focuses on how orbital speed varies in an elliptical orbit. A planet does not move at a constant speed; instead, its velocity changes depending on its distance from the Sun. The key idea comes from conservation of angular momentum and energy in orbital motion.

At the closest point (perihelion), the planet moves fastest, while at the farthest point (aphelion), it moves slowest. The ratio of speeds depends on how stretched the orbit is, which is measured by eccentricity. For small eccentricity values, the orbit is nearly circular, so the speed variation is small.

Using the relationship between distances and speeds, the ratio can be approximated as (1 + e)/(1 − e), where e is eccentricity. Since Earth’s eccentricity is very small, this ratio will be only slightly greater than 1.

This is similar to a skater pulling arms in to spin faster and extending them to slow down.

In summary, a nearly circular orbit leads to only a slight difference between maximum and minimum orbital speeds.

Explanation: The problem involves determining total energy in a system where a particle moves under a central force derived from a given potential energy function. Total energy is the sum of kinetic and potential energies.

For circular motion under a central force, the force is obtained from the negative gradient of potential energy. This force provides the necessary centripetal force for circular motion. By differentiating the potential energy function with respect to r, the force can be calculated.

Once the force is known, it is equated to the centripetal force expression involving velocity. This allows determination of kinetic energy. The total energy is then obtained by adding this kinetic energy to the given potential energy at that radius.

This is similar to balancing forces in circular motion where inward pull keeps the object moving in a fixed path.

In summary, energy in circular motion under a central force depends on both the form of potential energy and the balance with centripetal motion.

Explanation: This question is about identifying the physical principle behind a well-known observational law of planetary motion. Kepler’s second law states that a line joining a planet and the Sun sweeps out equal areas in equal intervals of time.

This behavior is directly linked to the conservation of angular momentum. When no external torque acts on a system, angular momentum remains constant. In planetary motion, the gravitational force acts along the line joining the planet and the Sun, producing no torque about the center.

Because angular momentum remains constant, the rate at which area is swept out also remains constant. This explains why planets move faster when closer to the Sun and slower when farther away.

An analogy is a spinning ice skater who changes speed without external torque.

In summary, equal area sweeping arises from the conservation of angular momentum in a torque-free central force system.

Explanation: Moment of inertia measures how Mass is distributed relative to an axis of rotation. It determines how difficult it is to change an object’s rotational motion. Objects with Mass concentrated closer to the axis have lower moment of inertia.

Different shapes distribute Mass differently. A ring has all its mass far from the axis, giving it a high moment of inertia. A Solid sphere has more mass closer to the center, resulting in a lower value. Cylinders fall somewhere in between depending on whether they are hollow or Solid.

The general idea is that spreading mass outward increases resistance to rotation. This is why compact shapes tend to rotate more easily than extended ones.

This is similar to holding a rod at the center versus at the ends—the distribution affects how hard it is to rotate.

In summary, the object with mass concentrated nearest to the axis has the smallest moment of inertia.

Explanation: The question explores how gravitational acceleration changes with distance from Earth’s center. The acceleration due to gravity is not constant at all heights; it depends on how far an object is from Earth.

The relationship is given by g ∝ 1/R², where R is the distance from Earth’s center. As altitude increases, this distance becomes larger, which reduces the gravitational pull experienced by the object.

Because of the inverse square dependence, even a moderate increase in distance leads to a noticeable decrease in gravitational acceleration. However, near the surface, this change is small and often approximated as constant.

This is similar to how the intensity of Light decreases as you move away from a source.

In summary, gravitational acceleration decreases with increasing altitude due to the inverse square dependence on distance.

Explanation: This problem involves symmetry in gravitational interactions. When identical masses are placed symmetrically, their forces can cancel out due to equal magnitude and opposite directions.

Each corner mass exerts a gravitational force on the central mass. The forces from opposite corners are equal in magnitude and directed along the diagonals but in opposite directions.

Because the square is perfectly symmetrical, the Vector sum of all these forces cancels out. This means the NET gravitational force at the center becomes zero.

This is similar to equal pulls from opposite directions balancing each other completely.

In summary, symmetry in mass distribution leads to cancellation of forces, resulting in zero NET gravitational force at the center.

Explanation: This question examines how weight changes when Earth’s size increases but its mass remains constant. Weight depends on gravitational acceleration, which is influenced by both mass and radius.

The formula for gravitational acceleration is g ∝ M/R². If the diameter doubles, the radius also doubles. Substituting this into the relation, the denominator becomes four times larger.

Since mass remains unchanged, the overall gravitational acceleration decreases significantly. Because weight is directly proportional to g, it also decreases accordingly.

This is like spreading the same amount of mass over a larger area, weakening the gravitational pull at the surface.

In summary, increasing Earth’s radius while keeping mass constant reduces surface gravity and thus decreases weight.

Explanation: The weight of an object on Earth depends not only on gravity but also on Earth’s rotation. At the equator, the rotational motion produces a centrifugal effect that slightly reduces the effective weight.

At the poles, this centrifugal effect is absent because the rotational axis passes through them. As a result, the only force acting is gravitational, making the effective weight slightly higher.

Additionally, Earth is slightly flattened at the poles, reducing the radius and increasing gravitational pull there.

This is similar to feeling lighter on a spinning ride compared to standing still.

In summary, weight increases slightly when moving from the equator to the poles due to reduced centrifugal effect and smaller radius.

Explanation: This question relates to a fundamental concept in Physics concerning mass. Inertial mass measures resistance to acceleration, while gravitational mass determines the strength of gravitational interaction.

Experiments have shown that these two types of mass are equivalent in magnitude. This equivalence is a cornerstone of classical mechanics and is also central to Einstein’s theory of general relativity.

Because of this equality, objects fall at the same rate in a gravitational field regardless of their mass, provided air resistance is negligible.

This idea is demonstrated in experiments like dropping different objects in a vacuum.

In summary, inertial and gravitational masses are equal, forming a key principle in Physics.

Explanation: This question asks how gravitational acceleration changes when the height above Earth’s surface equals its radius. The total distance from Earth’s center becomes twice the original radius.

Gravitational acceleration follows the inverse square law, meaning it varies as 1/R². When the distance doubles, the denominator becomes four times larger.

This significantly reduces the value of gravitational acceleration at that height compared to its surface value. The reduction is not linear but follows the square relationship.

This is similar to how intensity of forces decreases rapidly with distance in many physical systems.

In summary, doubling the distance from Earth’s center reduces gravitational acceleration by a factor determined by the inverse square law.

Explanation: This question explores how a pendulum behaves in a satellite that is freely orbiting Earth. The time period of a simple pendulum depends on gravitational acceleration, as given by T ∝ √(1/g). This means the restoring force responsible for Oscillation comes from gravity.

Inside an orbiting satellite, both the pendulum and the satellite are in continuous free fall toward Earth. This creates a condition known as apparent weightlessness, where effective gravitational acceleration inside the satellite becomes zero.

Since the restoring force vanishes, the pendulum bob does not return to its equilibrium position when displaced. Without this restoring force, oscillatory motion cannot occur, and the concept of a finite time period breaks down.

This is similar to astronauts floating in space where objects do not swing like they do on Earth.

In summary, in the absence of effective gravity inside a satellite, a simple pendulum cannot oscillate, so its time period becomes undefined.

Explanation: The motion of a satellite depends on the gravitational force provided by the planet, which acts as the centripetal force. This force is proportional to the planet’s mass, meaning any sudden change in mass affects the orbit.

If the planet’s mass is reduced abruptly, the gravitational pull weakens instantly. However, the satellite’s velocity at that moment remains unchanged because no immediate force alters it.

With reduced gravitational attraction, the existing velocity may no longer correspond to a stable circular orbit. The satellite now has more kinetic energy than required for the new gravitational condition, causing its path to change.

This situation is similar to reducing the tension in a string while an object is rotating—the object tends to move outward.

In summary, a sudden decrease in planetary mass disturbs orbital balance, leading the satellite to shift from its original orbit due to insufficient gravitational binding.

Explanation: This question deals with Fluid mechanics and pressure differences in a liquid. Air bubbles in water experience buoyant forces due to pressure variation with depth, as described by Fluid statics.

When two bubbles are close to each other, the pressure distribution around them becomes uneven. The region between the bubbles experiences slightly lower pressure compared to the outer sides.

This pressure difference causes a NET force pushing the bubbles toward each other. As a result, they tend to move closer and may even merge into a larger bubble.

This behavior is similar to how objects move toward regions of lower pressure in fluids.

In summary, unequal pressure distribution around nearby bubbles creates a NET force that brings them together.

Explanation: This question focuses on the mutual gravitational interaction between two bodies in space. According to Newton’s law of Gravitation, both bodies exert equal and opposite forces on each other.

Because of this mutual attraction, neither body remains completely stationary unless one mass is overwhelmingly larger. Instead, both objects revolve around a common point known as the center of mass.

The position of this center depends on the relative masses of the two bodies. If one is much heavier, the center lies closer to it, but motion still occurs for both.

This is similar to two people spinning while holding hands, both moving around a shared center.

In summary, two interacting celestial bodies orbit a common center of mass due to mutual gravitational forces.

Explanation: The time period of a satellite depends on its orbital radius according to Kepler’s third law, which states T ∝ R^3/2. This means even small changes in radius can affect the orbital period.

When the radius increases slightly, the period increases as well, but not in a linear manner. Instead, the increase follows a power relation, making the change somewhat moderate.

To estimate the percentage increase, one can use approximation techniques for small variations. A small fractional increase in radius results in a proportionally larger fractional increase in time period.

This is similar to how increasing the length of a circular path slightly increases the time taken to complete one round.

In summary, a small increase in orbital radius leads to a slightly larger percentage increase in time period due to the power dependence.

Explanation: This question examines which physical phenomena depend directly on gravitational force. Gravity influences weight, pressure in fluids, and buoyancy effects in liquids.

Archimedes’ buoyant force arises due to pressure differences in a Fluid, which are themselves caused by gravity acting on the Fluid. If gravity weakens, the pressure gradient in the Fluid decreases.

As a result, the buoyant force experienced by objects immersed in the Fluid also reduces. Other forces like viscous or electrostatic forces do not directly depend on gravity.

This is similar to how reduced gravity in space affects how objects float in liquids.

In summary, forces that rely on pressure differences caused by gravity are directly affected when gravitational strength changes.

Explanation: A geostationary satellite is one that appears stationary relative to a point on Earth’s surface. For this to happen, the satellite must rotate around Earth at the same rate as Earth itself rotates.

Earth completes one full rotation about its axis in a fixed amount of time, which determines the required orbital period for such satellites. Matching this rotation ensures the satellite stays above the same geographic location.

This synchronization is essential for Communication and weather satellites, as it allows continuous coverage over a specific region.

This is similar to two rotating systems moving in perfect synchrony so that their relative position remains unchanged.

In summary, the orbital period of a geostationary satellite must match Earth’s rotational period to maintain a fixed position relative to the surface.