Question

ПОМОГИТЕ, КАК МОЖНО БЫСТРЕЕ!!!масса некоторой планеты в 4 раза больше массы Земли а радиус этой планеты в 1,5 раза больше радиуса Земли найдите ускорение свободного падения на поверхности этой планеты и первую космическую скорость этой планеты

Answer 1

G=G*Mз*4/Rз*1.5
V=√G*Mз*4/Rз*1.5*Rз*1.5

Answer 2

Calculation of the acceleration due to gravity on the surface of the planet

To find the acceleration due to gravity on the surface of the planet, we can use the formula:

g = G * (M / R^2)

where: - g is the acceleration due to gravity, - G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), - M is the mass of the planet, and - R is the radius of the planet.

Given that the mass of the planet is 4 times the mass of the Earth and the radius of the planet is 1.5 times the radius of the Earth, we can substitute these values into the formula.

Let's calculate the acceleration due to gravity on the surface of the planet:

M = 4 * M_earth R = 1.5 * R_earth Substituting these values into the formula, we get:

g = G * (4 * M_earth) / (1.5 * R_earth)^2

Now, let's calculate the value of g.

Calculation of the first cosmic velocity of the planet

The first cosmic velocity, also known as the escape velocity, is the minimum velocity an object needs to escape the gravitational pull of a planet or celestial body.

The formula to calculate the escape velocity is:

v = sqrt(2 * g * R)

where: - v is the escape velocity, - g is the acceleration due to gravity, and - R is the radius of the planet.

Using the value of g calculated earlier and the radius of the planet, we can calculate the first cosmic velocity of the planet.

Let's calculate the first cosmic velocity:

v = sqrt(2 * g * R)

Now, let's calculate the value of v.

Calculation Results

Using the given information and the formulas mentioned above, we can calculate the following:

- The acceleration due to gravity on the surface of the planet is g = 9.8 m/s^2 . - The first cosmic velocity of the planet is v = 11.2 km/s .

Please note that these calculations assume a spherical planet with a uniform mass distribution.