(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
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(1a) The gravitational field strength g on the earth's surface is 9.8 N/kg. This means that for every kilogram of mass on the earth's surface, it experiences a force of 9.8 Newtons due to the gravitational pull of the earth. In other words, it is a measure of the intensity of the gravitational field at that location.
(b) According to the law of gravity, 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. Mathematically, we can express this as:
F = G * (m1 * m2) / r^2
Where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.
Now, let's consider an object of mass m near the earth's surface at a distance r from its center, where r > Re (the radius of the earth). The mass of the object is small compared to the mass of the earth, so we can assume m1 ≈ 0 and m2 ≈ M (mass of earth). Therefore, we can rewrite the equation as:
F = g * m = G * (0 * M) / r^2
Simplifying this equation, we find:
g = G * M / r^2
Rearranging, we get:
g * r^2 = G * M
This implies that the product of the gravitational field strength (g) and the square of the distance (r) from the center of the earth is equal to G (the gravitational constant) multiplied by the mass of the earth (M).