(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
LOL, I just did most of this in your problem below.
(1a) The gravitational field strength, g, on the Earth's surface being 9.8 N/kg means that every kilogram of mass on the Earth experiences a force of 9.8 Newtons (N) due to the Earth's gravity. In simpler terms, it indicates how strong the gravitational pull is on the surface of the Earth. This value of 9.8 N/kg is essentially equivalent to the acceleration due to gravity on the Earth's surface.
(b) Using the law of gravity, we can show that gr^2 = k, where g is the gravitational field strength at a distance r from the center of the Earth (r > Re) and Re is the radius of the Earth.
The law of gravity, formulated by Sir Isaac Newton, states that the force of gravity between two objects is directly proportional to the product of their masses (m1 and m2) and inversely proportional to the square of the distance (r) between their centers of mass. Mathematically, it can be expressed as:
F = G * (m1 * m2) / r^2
Where F is the gravitational force, G is the gravitational constant.
In this case, let's consider a mass m placed at a distance r from the center of the Earth. The force experienced by this mass due to Earth's gravity can be denoted as F = m * g, where g is the gravitational field strength at distance r.
Equating the two formulas, we have:
m * g = G * (m * M) / r^2
Where M is the mass of the Earth.
The mass m cancels out from both sides of the equation, and rearranging the formula, we obtain:
g * r^2 = G * M
Since G and M are constants, we can combine them into a new constant k, giving us:
gr^2 = k
This equation shows the relationship between the gravitational field strength and the distance from the center of the Earth.
(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 principle of conservation of energy. The initial potential energy (mgh) of the body, due to its height above the Earth's surface, will convert into kinetic energy (1/2mv^2) as it falls.
Let's consider the situation where the body is released from rest at a height h above the Earth's surface.
The initial potential energy is given by mgh, where m is the mass of the body, g is the gravitational field strength near the Earth's surface (approximately 9.8 N/kg), and h is the distance from the point of release to the surface of the Earth.
At the surface of the Earth, the potential energy is zero since the body has fallen to its lowest point. Thus, all of the initial potential energy is converted into kinetic energy.
Setting the initial potential energy equal to the final kinetic energy, we have:
mgh = (1/2)mv^2
Cancelling out the mass, we get:
gh = (1/2)v^2
Simplifying further:
v^2 = 2gh
Taking the square root of both sides:
v = √(2gh)
This equation allows us to calculate the velocity of the body when it strikes the Earth's surface.
(3a) To calculate the energy required to transfer a 500 kg spaceship from a circular orbit of radius 2Re to a circular orbit of radius 4Re around the Earth, we need to consider the change in potential energy.
The potential energy of an object in orbit is given by the formula:
U = -G * (M * m) / R
Where U is the potential energy, G is the gravitational constant, M is the mass of the Earth, m is the mass of the spaceship, and R is the radius of the orbit.
Let's denote the initial potential energy as U1 when the spaceship is in a circular orbit of radius 2Re, and the final potential energy as U2 when the spaceship is in a circular orbit of radius 4Re.
The energy required to transfer the spaceship is then given by the difference between the final and initial potential energies:
ΔU = U2 - U1
Substituting the appropriate values into the formula, we have:
ΔU = -G * (M * m) / (4Re) - (-G * (M * m) / (2Re))
Simplifying the expression, we get:
ΔU = G * (M * m) * (1 / (2Re) - 1 / (4Re))
ΔU = G * (M * m) * (2 / (8Re) - 1 / (4Re))
ΔU = G * (M * m) * (1 / (8Re))
(3b) Discussing the change in potential energy, kinetic energy, and the total energy of the system during the transfer:
- Potential Energy: As calculated in part (3a), the potential energy of the spaceship increases as it moves from a circular orbit of radius 2Re to a circular orbit of radius 4Re. This increase in potential energy is due to the spaceship being farther from the center of the Earth, resulting in a larger gravitational force and thus a greater potential energy.
- Kinetic Energy: The kinetic energy of the spaceship remains unchanged during the transfer. In a circular orbit, the kinetic energy is given by the formula:
KE = (1/2)mv^2
Since the velocity of the spaceship remains constant, the kinetic energy doesn't change.
- Total Energy: The total energy of the spaceship system, represented by the sum of potential energy and kinetic energy, remains constant during the transfer. This is known as the conservation of mechanical energy. Any increase in potential energy is offset by a decrease in kinetic energy and vice versa.
Therefore, during the transfer, the potential energy increases, but the kinetic energy remains the same, resulting in a constant total energy for the system.