Final answer:
The time period of a pendulum on a planet with twice the mass and diameter of Earth would be √2 times longer than on Earth, as gravitational field strength is inversely proportional to the square of the radius and directly proportional to the mass.
Step-by-step explanation:
The time period of a pendulum is given by the formula T = 2π√(L/g), where T is the time period, L is the length of the pendulum, and g is the acceleration due to gravity. On Earth, a seconds pendulum has a period of 2 seconds, and the acceleration due to gravity is approximately 9.81 m/s². On a planet with twice the mass and diameter of Earth, the gravitational field would be affected. The gravitational field strength, g, is calculated by the formula g = G*M/R², where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.
For a planet with twice the mass (2M) and twice the diameter (hence, twice the radius 2R) of Earth, the gravitational field would be g = G*(2M)/(2R)² = (1/2)*G*M/R², which is half of Earth's gravitational field. This means that the new acceleration due to gravity would be approximately 4.905 m/s². Consequently, the time period of the pendulum would increase, as it is inversely proportional to the square root of the acceleration due to gravity. Applying this to the formula, the new time period T' would be 2π√(L/(1/2)g₀), where g₀ is the acceleration due to gravity on Earth. So, the time period of the pendulum on this new planet would be 2π√(2L/g₀) = 2√2*T, which means it would be √2 times longer than on Earth.