Gravitation

The mass of a satellite does not affect the time period

T_A= T_B = 1:2√2

Hence, the correct option is d) 1:2√2.

The correct option is (C) :

Change in G.P.E. = final energy – initial energy
= =

Gravitational force does not depend on the medium.
In this case time of flight of a ball ≥ 2 × 2 = 4 sec.
∴ Time of flight = 2u/g ≥4
u ≥ 2g u ≥ 19.6 m/s (∴g = 9.8 m/s²)

The dimensions of the universal gravitational constant are M^–1L³T^–2.

Gravitational Potential and Height Calculation

The gravitational potential at a distance is given by the formula:

....(1)

The gravitational field strength at the same distance is given by the formula:

......(2)

Dividing Equations (1) and (2):

By dividing equation (1) by equation (2), we get:

Solving for :

Therefore, the height is:

Given that the radius of the Earth is approximately 6400 km, we have:

Conclusion:

The height is approximately .

Gravitational Potential Energy and Gravitational Field

We are working with the gravitational potential energy and gravitational field for a mass at a distance from the center of a planet with mass .

Step 1: Gravitational Potential Energy

The gravitational potential energy of a mass in a gravitational field is given by the formula:

Where: - is the gravitational constant, - is the mass of the planet, - is the mass of the object, - is the distance from the center of the planet to the object.

Step 2: Gravitational Field at the Surface

The gravitational field strength at the surface of the planet is given by the formula:

Where is the radius of the planet, and is the gravitational field strength at the surface.

Step 3: Relating and

Rearranging the equation for , we get the relationship:

Step 4: Gravitational Potential Energy for Height

If the object is at a height above the surface, the gravitational potential energy becomes:

Where: - is the radius of the planet, - is the height above the surface of the planet, - is the mass of the object.

Conclusion:

The gravitational potential energy at a height above the surface is given by:

The two astronauts will more towards each due to the gravitational force b/w them but it will be a very slow process due to their law mass.
Therefore, the correct option is (A) : Move towards each other.

As per, Kepler's Law - = constant

As dA/dt is constant, point A is perihelion and C is aphelion.

So, K_A > K_B > K_C

Read More: Gravitation

To solve this problem, we need to determine the maximum height reached by a particle projected upwards from the Earth's surface with an initial velocity defined in terms of escape velocity. Let's break down the solution step-by-step:

The given data includes:

• Mass of the particle
• Escape velocity
• Initial velocity where

The escape velocity from the Earth's surface is given by:

where:

• is the gravitational constant
• is the mass of the Earth
• is the radius of the Earth

The initial kinetic energy of the particle is:

The initial gravitational potential energy at the Earth's surface is:

At the maximum height above the Earth's surface, the particle will momentarily stop, and all initial kinetic energy is converted into gravitational potential energy.

Equating the initial total energy to the total energy at the maximum height, we have:

From the expression of escape velocity, substitute to get:

Simplifying this equation:

Further simplification gives:

This can be simplified to:

Cancelling GMm from both sides and solving for , we get:

Hence, the maximum height above the Earth's surface is:

Therefore, the correct answer is:

• F is the gravitational force experienced by the body, which is 3.0 N.
• m is the mass of the body. Here, the mass is given in grams, so we need to convert it to kilograms. Since 1 kg = 1000 g, the mass is .

To find the point where the gravitational field equals zero between two masses, we can use the principle of superposition. Let the two masses be and , placed at a distance apart. Let the gravitational field be zero at a point at a distance from mass . The distance from mass then becomes .

The gravitational field due to a mass at a distance is given by:

For the gravitational fields to cancel out:

Simplifying, we get:

Taking the square root on both sides:

Cross-multiplying gives:

Solve for :

Now, calculate the gravitational potential at this point:

The potential due to a mass at a distance is:

The total potential is the sum of the potentials due to both masses:

Potential due to mass at distance :

Potential due to mass at distance :

Total potential:

Thus, the correct answer is:

To determine how the quantity relates to the satellite's orbital period , we begin with relevant physics principles. The satellite is orbiting just above the Earth's surface, implying it is in a low Earth orbit. The gravitational force is the centripetal force required to keep the satellite in orbit. Using the formula for gravitational force and centripetal force , equating both gives:

Cancel and from both sides:

The relationship between velocity and orbital period is , substituting gives:

Simplify:

Rearranging terms leads to:

For a satellite just above Earth’s surface, is approximately equal to Earth's radius , thus:

Given and comparing it with the derived , it suggests that is proportional to , confirming the relationship. Therefore, the correct answer is .