It is not space that is curved, it is spacetime that is curved. You are neglecting the time part of spacetime and drawing the wrong conclusions.binis said:
No. Different speeds are different directions in spacetime and the curvature of spacetime can be different in different directions. This is essentially why the Newtonian gravitational deflection for light is off by a factor of 2 from the relativistic deflection.binis said:
No, that is not necessary. The spatial direction is only relevant to avoid colliding with the earth.binis said:
How is it calculated by GR? Please provide a link.Ibix said:
What about the last query? " In that case, since gravity is not a force, no escape velocity is required?"Dale said:
Chapter 7 of Carroll's online lecture notes on GR. From memory, equations 7.43, 7.44, 7.47 and 7.48 are the relevant ones; you'll need to set ##\epsilon=1## for a timelike particle, ##E=1## for a "stop at infinity" case, and calculate the magnitude of the three velocity measured by an observer hovering at ##r=R_E##.binis said:
It makes no sense. Assuming there's a way to define "arriving at infinity" you can always ask what velocity achieves it. In the Schwarzschild case (in my last post) it's even done in the background by a conservation of energy argument.binis said:
It is a non sequitur, which is a form of logical fallacy. Gravity not being a force in no way implies that there is no escape velocity required.binis said:
You have to consider space-time, not just space.binis said:
Earth's escape velocity is the minimum speed that an object must reach to break free from Earth's gravitational pull without further propulsion. This velocity is approximately 11.2 kilometers per second (about 25,000 miles per hour) at the surface of the Earth.
Earth's escape velocity is calculated using the formula \( v = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2), \( M \) is the mass of the Earth (approximately 5.972 × 10^24 kg), and \( R \) is the radius of the Earth (about 6,371 kilometers). Plugging in these values gives the escape velocity of approximately 11.2 km/s.
Yes, Earth's escape velocity decreases with altitude. As you move further away from the Earth's surface, the gravitational pull weakens, and thus a lower velocity is required to escape Earth's gravity. The escape velocity at any given altitude can be calculated using the same formula, but with the radius \( R \) adjusted to include the altitude above Earth's surface.
Yes, Earth's escape velocity is the same for all objects regardless of their mass. The escape velocity depends only on the gravitational constant, the mass of the Earth, and the distance from the center of the Earth. Therefore, whether it's a small satellite or a large spacecraft, the escape velocity remains 11.2 km/s at the Earth's surface.
Earth's escape velocity is moderate compared to other planets in our solar system. For instance, Mercury has a lower escape velocity of about 4.3 km/s due to its smaller mass and size, while Jupiter has a much higher escape velocity of around 59.5 km/s because of its massive size and strong gravitational pull. Each planet's escape velocity is determined by its mass and radius.