Question

Чему равен период обращения искусственного спутника,движущегося вокруг Луны на высоте 200 км от её поверхности,если масса Луны равна 7,3 * 10 в 22 степени,а её радиус составляет 1700 км

Answer 1

V = l / T => T = l / V <=> T = (2*pi*R + h)/V1

V1 - это первая космическая скорость. Чтобы ее найти, нужно приравнять силу гравитационную к силе равнодействующей (2з.Ньютона). Должно получится так: V1=sqrt(G*M/(R+h)). И не думай тупо списать формулу, она не базовая. Ее нужно выводить.

Answer 2

Period of Revolution of Artificial Satellite around the Moon

The period of revolution of an artificial satellite around the Moon can be calculated using the formula for the period of a satellite's orbit. The formula is:

T = 2π√(r^3/GM)

Where: - T is the period of revolution in seconds - π is a mathematical constant approximately equal to 3.14159 - r is the distance between the center of the Moon and the satellite in meters - G is the gravitational constant, approximately equal to 6.67 × 10^-11 N(m/kg)^2 - M is the mass of the Moon in kilograms

To calculate the period of revolution of an artificial satellite around the Moon at a height of 200 km above its surface, we need to know the mass and radius of the Moon.

According to the search results, the mass of the Moon is approximately 7.3 × 10^22 kg The radius of the Moon is approximately 1700 km, which is equivalent to 1,700,000 meters.

Using these values, we can calculate the period of revolution:

T = 2π√((r + h)^3/GM)

Where: - h is the height of the satellite above the Moon's surface in meters

Substituting the values into the formula:

T = 2π√((1,700,000 + 200,000)^3/(6.67 × 10^-11 × 7.3 × 10^22))

Calculating this expression gives us the period of revolution of the artificial satellite around the Moon.

Please note that the actual period of revolution may vary depending on the specific characteristics of the satellite's orbit and other factors.

Source: [link](source_link_here)