please help solve these
questions
(1a). The gravitational field strength
g on the earth’s surface is
9.8N/kg. explain what this
means.(b) Using the law of gravity,
show that gr2= k. Where g=gravitational field strength at a
distance r from the center of the
earth (r>Re), where Re is the radius of the earth.
(2). Determine the velocity which a body released at a
distance r from the center of the
earth when it strikes the earth surface.
(3). A 500kg spaceship is in a circular orbit of radius 2Re about
the earth. (a)How much energy is required to transfer the spaceship to a
circular orbit of radius 4Re. (b)Discuss the change in potential
energy, kinetic energy and the
total energy of the system.
(4). State the universal law of gravitation. (b)Derive the
relationship between the gravitational constant G and
acceleration of free fall g at a point close to the earth surface
(assume that the earth is a sphere of uniform density).
(5). Show that theoretically the
period of rotation of a satalite T,
which circles a planet at negligible distance from its surface depends only on the
density of the planet
(1a) The gravitational field strength, denoted by g, on the Earth's surface is 9.8 N/kg. This means that for every kilogram of mass on the Earth's surface, there is a force of 9.8 Newtons pulling it towards the center of the Earth. It represents the acceleration due to gravity experienced by objects on Earth.
(b) To show that gr^2 = k using the law of gravity, we first need to understand the law of gravity, which states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Let's consider a small object of mass m at a distance r from the center of the Earth. The force of gravity acting on this object can be expressed as F = mg, where g is the gravitational field strength at a distance r from the center of the Earth.
Now, let's consider an object at the Earth's surface (r = Re, where Re is the radius of the Earth). The force of gravity acting on this object is given by F = m * g, from the equation above.
Using the law of gravity, we can equate these two expressions for the force:
m * g = G * (m * M) / R^2
where G is the gravitational constant and M is the mass of the Earth, and R is the radius of the Earth.
Simplifying the equation, we get:
g = G * M / R^2
Now, let's consider the gravitational field strength at a distance r from the center of the Earth.
g = G * M / r^2
We can rearrange this equation to get:
g * r^2 = G * M
Since G * M is a constant, let's call it k.
Therefore, gr^2 = k.
(2) To determine the velocity at which a body, released at a distance r from the center of the Earth, strikes the Earth's surface, we can use the concept of conservation of mechanical energy.
At the starting point, the body has potential energy due to its position, given by U = -G * (m * M) / r, where m is the mass of the body and M is the mass of the Earth.
As the body falls towards the Earth's surface, it loses potential energy and gains an equal amount of kinetic energy, given by K = (1/2) * m * v^2, where v is the velocity of the body.
At the point of impact, the potential energy becomes zero, and the total mechanical energy of the system is given entirely by the kinetic energy.
Therefore, we can equate the initial potential energy to the final kinetic energy:
-G * (m * M) / r = (1/2) * m * v^2
Simplifying the equation, we get:
v = sqrt(2 * G * M / r)
So, the velocity at which the body strikes the Earth's surface is given by v = sqrt(2 * G * M / r).
(3) (a) To determine the energy required to transfer the spaceship from a circular orbit of radius 2Re to a circular orbit of radius 4Re, we can use the concept of conservation of mechanical energy.
The initial mechanical energy in the 2Re orbit is given by E1 = -(G * m * M) / (2Re), where m is the mass of the spaceship and M is the mass of the Earth. The negative sign indicates that the energy is in the form of potential energy.
The final mechanical energy in the 4Re orbit is given by E2 = -(G * m * M) / (4Re).
The energy required to transfer the spaceship from the initial orbit to the final orbit is the difference between these energies:
Energy = E2 - E1 = -(G * m * M) / (4Re) + (G * m * M) / (2Re)
Simplifying the equation, we get:
Energy = (G * m * M) / (4Re) - (G * m * M) / (2Re)
(b) The change in potential energy, kinetic energy, and the total energy of the system can be determined.
In the initial orbit (2Re), the potential energy is higher due to the larger distance from the center of the Earth. As the spaceship moves to the final orbit (4Re), the potential energy decreases.
The kinetic energy of the spaceship remains the same, as the mass is constant throughout the transfer between orbits.
The total energy of the system, which is the sum of potential energy and kinetic energy, remains constant during the transfer.
(4) The universal law of gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematically, the law of gravitation can be represented as:
F = G * (m1 * m2) / r^2
where F is the force of gravity between two objects, m1 and m2 are the masses of the two objects, r is the distance between their centers, and G is the gravitational constant.
To derive the relationship between the gravitational constant G and the acceleration of free fall g at a point close to the Earth's surface, we can consider a small object of mass m placed at a distance r from the center of the Earth, where r is greater than the radius of the Earth (Re).
Using the law of gravitation, the force of gravity acting on the object can be expressed as:
F = G * (m * M) / r^2
where M is the mass of the Earth.
Now, considering that the object is close to the Earth's surface, we can approximate the distance r to be very close to the radius of the Earth (r ≈ Re).
Therefore, the force of gravity acting on the object can be approximated as:
F = G * (m * M) / Re^2
The object will experience a gravitational force that is equal to its weight, given by F = m * g, where g is the acceleration due to gravity.
Equating these two expressions for the force, we get:
m * g = G * (m * M) / Re^2
Simplifying the equation, we get:
g = G * M / Re^2
This equation shows the relationship between the gravitational constant G and the acceleration of free fall g at a point close to the Earth's surface.
(5) The period of rotation of a satellite, denoted by T, which circles a planet at a negligible distance from its surface depends only on the density of the planet.
To understand this, consider a satellite in a circular orbit around a planet at a distance very close to its surface. The gravitational force acting on the satellite provides the centripetal force required to keep it in orbit.
The centripetal force acting on the satellite can be expressed as:
F_c = m * (v^2 / r)
where m is the mass of the satellite, v is its orbital velocity, and r is the radius of its orbit (which is negligible compared to the radius of the planet).
The gravitational force acting on the satellite is given by:
F_g = G * (m * M) / (r^2)
where M is the mass of the planet.
Since the gravitational force provides the centripetal force, we can equate the two expressions for the force:
G * (m * M) / (r^2) = m * (v^2 / r)
Simplifying the equation, we get:
(v^2 / r) = G * M / (r^2)
The orbital velocity v can be calculated as:
v = sqrt(G * M / r)
The time taken for one complete rotation (the period T) can be determined by dividing the circumference of the orbit by the orbital velocity:
T = (2πr) / v
Substituting the value of v from the previous equation, we get:
T = (2πr) / sqrt(G * M / r)
Simplifying the equation, we find:
T = 2π * sqrt(r^3 / (G * M))
Therefore, the period of rotation of a satellite depends only on the density of the planet, as it is independent of the radius of orbit and the mass of the satellite.