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what will be the time interval measured by a clock at rest on the surface of a neutron star of radius 10 km with schwarzschild radius 3km, when a clock at rest at a very large distance measures a time interval 1hr

User Matthewb
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Final answer:

The time interval measured by a clock at rest on the surface of a neutron star can be calculated using the principle of time dilation. According to Einstein's theory of relativity, time moves slower in a strong gravitational field. The formula for time dilation is given by DeltaT' = DeltaT * sqrt(1 - 2GM/(c^2r)). By substituting the given values into the formula, the time interval measured by a clock at rest on the surface of the neutron star is approximately 3276.02 seconds.

Step-by-step explanation:

The time interval measured by a clock at rest on the surface of a neutron star can be calculated using the principle of time dilation. According to Einstein's theory of relativity, time moves slower in a strong gravitational field. The formula for time dilation is given by:

Δt' = Δt × √(1 - 2GM/(c₂r))

Where Δt is the time interval measured at a large distance from the neutron star, G is the gravitational constant, M is the mass of the neutron star, c is the speed of light, and r is the radius of the neutron star.

In this scenario, the time interval measured by a clock at rest on the surface of the neutron star can be calculated by substituting the given values into the formula. Let's calculate:

Given:

Δt = 1 hour = 3600 seconds

r = 10 km = 10,000 meters

Schwarzschild radius (Rs) = 3 km = 3,000 meters

Plugging these values into the formula:

Δt' = 3600 × √(1 - 2GM/(c₂r))

Δt' = 3600 × √(1 - 2(6.67430 × 10⁻⁻11 N m₉2 kg⁻⁵2)(2.0 × 10⁻ⁱ30 kg)/((3.0 × 10⁻⁴ m/s)₂(10,000 m)))

Δt' = 3600 × √(1 - 2(1.33536 × 10⁻⁰1)(6.67 × 10⁻ⁱ3)/(3.0 × 10⁻⁹))

Δt' = 3600 × √(1 - 2(0.08472))

Δt' = 3600 × √(1 - 0.16944)

Δt' = 3600 × √(0.83056)

Δt' = 3600 × 0.91001

Δt' = 3276.02 seconds

Therefore, the time interval measured by a clock at rest on the surface of the neutron star would be approximately 3276.02 seconds.

User Tammy
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