The International Space Station generally operates in orbits at altitudes between 330-435 km above the Earth's surface. You have probably seen astronauts float about "weightless" up there in "zero-g."
1) If the Earth's gravity was zero at that altitude, what does Newton's First Law tell you about the shuttle's movement?
2) What is the ratio of the acceleration due to the Earth's gravity g on the surface of the Earth to g at 400 km above the Earth (RE = 6400 km).
1) If the Earth's gravity was zero at that altitude, according to Newton's First Law, the shuttle would continue to move in a straight line at a constant velocity. In other words, it would zoom away from Earth like a duck escaping a poorly cooked dinner!
2) To find the ratio of the acceleration due to Earth's gravity at the surface to the gravity at 400 km above the Earth, we need a little math. The acceleration due to gravity is given by the formula g = GM/R^2, where G is the gravitational constant, M is the mass of the Earth, and R is the distance from the center of the Earth.
At the Earth's surface (R = RE = 6400 km), we have g1 = GM/RE^2. At 400 km above the Earth (R = RE + 400 km), we have g2 = GM/(RE + 400 km)^2.
So, the ratio g1/g2 = (GM/RE^2) / (GM/(RE + 400 km)^2). Let's simplify this:
g1/g2 = (RE + 400 km)^2 / RE^2
Plugging in the values, we have:
g1/g2 = (6400 km + 400 km)^2 / (6400 km)^2
Simplifying further, we get:
g1/g2 = (6800 km)^2 / (6400 km)^2
And finally,
g1/g2 ≈ 1.129
So, the ratio of the acceleration due to Earth's gravity at the surface to that at 400 km above is approximately 1.129. Just imagine, if gravity was like comedy, it would have a refreshingly low altitude of jokes per second.
1) Newton's First Law of Motion, also known as the Law of Inertia, states that an object at rest will stay at rest, and an object in motion will continue moving at a constant velocity in a straight line, unless acted upon by an external force. If the Earth's gravity was zero at the altitude of the International Space Station, the shuttle would continue moving at a constant velocity in a straight line, maintaining its momentum unless influenced by other forces such as a thruster or gravitational pull from other celestial bodies.
2) The acceleration due to gravity, denoted as "g," is inversely proportional to the square of the distance between the object and the center of the Earth. So the ratio of g at the surface of the Earth to g at 400 km above the Earth can be calculated using the formula:
g_surface / g_400km = (RE / (RE + 400))^2
Where RE is the radius of the Earth (6400 km).
Plugging in the values:
g_surface / g_400km = (6400 / (6400 + 400))^2
= (6400 / 6800)^2
= (0.9412)^2
≈ 0.886
Therefore, the ratio of the acceleration due to the Earth's gravity on the surface of the Earth to the acceleration due to gravity at 400 km above the Earth is approximately 0.886.
To answer these questions, let's break them down one by one:
1) If the Earth's gravity was zero at that altitude, what does Newton's First Law tell you about the shuttle's movement?
Newton's First Law, also known as the law of inertia, states that an object in motion will remain in motion with a constant velocity unless acted upon by an external force. This means that if the Earth's gravity did not exist at the altitude of the International Space Station (ISS), the shuttle would continue to move with its current velocity in a straight line. The absence of gravity would eliminate the force that pulls objects towards the Earth, allowing the shuttle to float in a "weightless" or "zero-g" environment.
2) What is the ratio of the acceleration due to the Earth's gravity (g) on the surface of the Earth to g at 400 km above the Earth (RE = 6400 km)?
To determine the ratio of the acceleration due to gravity at these two locations, we can use the inverse square law. According to this law, the gravitational force between two objects is inversely proportional to the square of the distance between their centers.
The formula for the acceleration due to gravity, g, can be expressed as:
g = G * (M / R^2)
Where G is the gravitational constant (approximately 6.67430 x 10^(-11) N(m/kg)^2), M is the mass of the Earth (approximately 5.972 × 10^24 kg), and R is the distance between the center of the Earth and the object.
At the surface of the Earth (R = RE), the acceleration due to gravity is g_surface:
g_surface = G * (M / RE^2)
At a height of 400 km above the Earth's surface (R = RE + 400 km), the acceleration due to gravity is g_400:
g_400 = G * (M / (RE + 400 km)^2)
To find the ratio of g_surface to g_400, we can divide these two equations:
ratio = g_surface / g_400
Plugging in the values, the equation becomes:
ratio = (G * (M / RE^2)) / (G * (M / (RE + 400 km)^2))
The gravitational constant and the mass of the Earth (G and M) are constant, so they cancel out, leaving us with:
ratio = (1 / RE^2) / (1 / (RE + 400 km)^2)
Simplifying further:
ratio = (RE + 400 km)^2 / RE^2
Substituting the values for RE = 6400 km and simplifying the equation:
ratio = (6400 km + 400 km)^2 / (6400 km)^2
ratio = (6800 km)^2 / (6400 km)^2
ratio = 4624 / 4096
Therefore, the ratio of the acceleration due to gravity on the Earth's surface to the acceleration due to gravity at 400 km above the Earth is approximately 1.13.