A simple pendulum has a period T on the surface of planet Earth. When taken to the surface of planet X, the same pendulum has a period 2T.
a) If the radius of planet X is 1/2 the radius of Earth, what is the mass of Planet X, relative to the mass of Earth?
b) Suppose a vertical mass and spring oscillator has a period T on Earth. What period would the oscillator have on planet X? Explain.
a) If the period on planet X is 2T, the acceleration of gravity (g) at the surface is 4 times less than Earth's, because
Period = 2 pi sqrt(L/g)
Since g = GM/R^2, if g is 1/4 as large as Earth's, and R is half the Earth's radius, M is 1/16 of Earth's.
g(X) = g(earth)*[M(X)/M(earth)]*[Rearth/RX]^2
= g(earth)*(1/16)*2^2 = g(earth)/4
b) The period of a spring-mass oscillator system would not change. It does not depend upon gravity.
To find the answers to these questions, we need to use the formula for the period of a simple pendulum:
T = 2π√(L/g)
Where:
T = Period of the pendulum
π = Pi (approximately 3.14159)
L = Length of the pendulum
g = Acceleration due to gravity
Let's start with question a:
a) If the radius of planet X is 1/2 the radius of Earth, what is the mass of Planet X, relative to the mass of Earth?
By comparing the period of the pendulum on Earth (T) with the period of the pendulum on planet X (2T), we can equate the two periods using the formula:
T₁ = 2T₂
Substituting the formula for the period of the pendulum:
2π√(L₁/g₁) = 2π√(L₂/g₂)
Since we are comparing the two planets, the length of the pendulum (L) can be assumed to remain the same on both planets. Therefore, we can cancel out the L₁ and L₂ terms:
√(g₂/g₁) = 2
Squaring both sides of the equation, we get:
g₂/g₁ = 4
Now, let's break down the concept of gravity between two planets. The acceleration due to gravity (g) on a planet can be expressed as:
g = (G * M) / r²
Where:
G = Universal Gravitational Constant
M = Mass of the planet
r = Radius of the planet
Since the radius of planet X is 1/2 the radius of Earth, we can express the mass of planet X (M₂) relative to the mass of Earth (M₁) as:
M₂ = (r₂/r₁)² * M₁
Substituting these values into the earlier equation:
g₂/g₁ = [(G * (r₂/r₁)² * M₁) / r₂²] / [(G * M₁) / r₁²]
Simplifying, we get:
g₂/g₁ = (r₁/r₂)²
Substituting the value of g₂/g₁ from earlier, we have:
(r₁/r₂)² = 4
Taking the square root of both sides, we get:
r₁/r₂ = 2
Since the radius of planet X is 1/2 the radius of Earth, we can deduce that:
2 * r₂ = r₁
Simplifying, we find:
r₂ = r₁/2
Substituting this value into the equation for the mass of planet X (M₂):
M₂ = [(r₁/2)/r₁]² * M₁
Simplifying, we get:
M₂ = 1/4 * M₁
Therefore, the mass of Planet X is 1/4 the mass of Earth.
b) Suppose a vertical mass and spring oscillator has a period T on Earth. What period would the oscillator have on planet X? Explain.
The period of a vertical mass and spring oscillator is also given by:
T = 2π√(m/k)
Where:
T = Period of the oscillator
π = Pi (approximately 3.14159)
m = Mass of the object attached to the spring
k = Spring constant
In this case, we want to compare the periods of the oscillator on Earth (T₁) and planet X (T₂). We can equate the two periods using the formula:
T₁ = T₂
Substituting the formula for the period of the oscillator:
2π√(m₁/k₁) = 2π√(m₂/k₂)
Since we are comparing the two planets, the mass of the object attached to the spring (m) can be assumed to remain the same on both planets. Therefore, we can cancel out the m₁ and m₂ terms:
√(k₂/k₁) = 1
Squaring both sides of the equation, we get:
k₂/k₁ = 1
This implies that the spring constant (k) remains the same on both planets.
Therefore, the oscillator would have the same period (T) on planet X as it does on Earth.