Question

1. На какой высоте над поверхностью Земли необходимо запустить спутник со скоростью 7800 м/с, чтобы он смог вращаться вокруг планеты? Масса Земли 6·1024 кг, радиус Земли 6400 км. 2. вычислить высоту спутника пренебрежимо малой по сравнению с радиусом планеты.найдите первую космическую скорость для планеты радиусом 6670 км, имеющей массу 2·1024 кг. 3. Определите ускорение свободного падения для планеты, имеющей радиус 5000 км, если первая космическая скорость для нее равна 7 км/с.

Answer 1

Ответ:

35 786 км над уровнем моря

Answer 2

I will answer your questions in detail.

1. To launch a satellite with a speed of 7800 m/s at a height above the Earth's surface where it can orbit the planet, we need to use the formula for the circular motion of a satellite:

$$v = \sqrt{\frac{GM}{R+h}}$$

where $v$ is the speed of the satellite, $G$ is the gravitational constant, $M$ is the mass of the Earth, $R$ is the radius of the Earth, and $h$ is the height of the satellite above the surface. Solving for $h$, we get:

$$h = \frac{GM}{v^2} - R$$

Plugging in the given values, we get:

$$h = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{(7800)^2} - 6400 \times 10^3$$

$$h \approx 1.2 \times 10^6 \text{ m}$$

Therefore, the satellite should be launched at a height of about 1.2 million meters (or 1200 km) above the Earth's surface.

2. To find the first cosmic speed for a planet with a radius of 6670 km and a mass of 2×10^24 kg, we need to use the same formula as above, but set $h$ to zero, since we assume the satellite is at the surface of the planet. Therefore, we get:

$$v = \sqrt{\frac{GM}{R}}$$

Plugging in the given values, we get:

$$v = \sqrt{\frac{6.67 \times 10^{-11} \times 2 \times 10^{24}}{6670 \times 10^3}}$$

$$v \approx 5.6 \times 10^3 \text{ m/s}$$

Therefore, the first cosmic speed for this planet is about 5.6 km/s.

3. To determine the acceleration of free fall for a planet with a radius of 5000 km and a first cosmic speed of 7 km/s, we need to use the inverse of the formula for the first cosmic speed, and solve for $g$, the acceleration of free fall. Therefore, we get:

$$g = \frac{v^2}{R}$$

Plugging in the given values, we get:

$$g = \frac{(7000)^2}{5000 \times 10^3}$$

$$g \approx 9.8 \text{ m/s}^2$$

Therefore, the acceleration of free fall for this planet is about 9.8 m/s^2, which is very close to the value on Earth.как решать задачи по механике спутников