Help please i need an answer asap

When a planet is at its slowest orbital speed, its radius vector sweeps an area, A, in 45 days. What area will the radius vector

MArishka [77]

The area the radius vector will sweep is 0.889A

According to Kepler's second law, the radius vector sweeps out equal areas in equal times.

Let A = area and t = time period,

According to Kepler's law, A/t = constant

So, A₁/t₁ = A₂/t₂ where A₁ = area the radius vector sweeps at slowest orbital speed = A, t₁ = time period at slowest orbital speed = 45 days, A₂ = area the radius vector sweeps at fastest orbital speed, t₂ = time period at fastest orbital speed = 40 days.

Making A₂ subject of the formula, we have

A₂ = A₁t₂/t₁

Substituting the values of the variables into the equation, we have

A₂ = A₁t₂/t₁

A₂ = A × 40 days/45 days

A₂ = A × 40/45

A₂ = A × 8/9

A₂ = A × 0.889

A₂ = 0.889A

So, the area the radius vector will sweep is 0.889A

learn more about Kepler's second law here:

brainly.com/question/4639131

8 0

2 years ago

A solid ball of radius rb has a uniform charge density ρ.

dalvyx [7]

A) $E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}$

In this problem we have spherical symmetry, so we can apply Gauss theorem to find the magnitude of the electric field:

$\int E(r) \cdot dr = \frac{q}{\epsilon_0}$

where the term on the left is the flux of the electric field through the gaussian surface, and q is the charge contained in the surface.

Here we are analyzing the field at a distance $r>r_B$ , so outside the solid ball. If we take a gaussian sphere with radius r, we can rewrite the equation above as:

$E(r) \cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$ (1)

where $4 \pi r^2$ is the surface of the sphere.

The charge contained in the sphere, q, is equal to the charge density $\rho$ times the volume of the solid ball, $\frac{4}{3}\pi r_b^3$ :

$q= \rho (\frac{4}{3}\pi r_b^3)$ (2)

Combining (1) and (2), we find

$E(r) \cdot 4 \pi r^2 = \frac{4\rho \pi r_b^3}{3 \epsilon_0}\\E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}$

And we see that the electric field strength is inversely proportional to the square of the distance, r.

B) $\frac{\rho r}{3 \epsilon_0}$

Now we are inside the solid ball: $r$ . By taking a gaussian sphere with radius r, the Gauss theorem becomes

$E(r) \cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$ (1)

But this time, the charge q is only the charge inside the gaussian sphere of radius r, so

$q= \rho (\frac{4}{3}\pi r^3)$ (2)

Combining (1) and (2), we find

$E(r) \cdot 4 \pi r^2 = \frac{4\rho \pi r^3}{3 \epsilon_0}\\E(r) = \frac{\rho r}{3 \epsilon_0}$

And we see that this time the electric field strength is proportional to r.

C)

E(0)=0.

limr→∞E(r)=0.

The maximum electric field occurs when r=rb.

Explanation:

From part A) and B), we observed that

- The electric field inside the solid ball ( $r$ ) is

$\frac{\rho r}{3 \epsilon_0}$ (1)

so it increases linearly with r

- The electric field outside the solid ball ( $r>r_B$ ) is

$E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}$ (2)

so it decreases quadratically with r

--> This implies that:

1) At r=0, the electric field is 0, because if we substitute r=0 inside eq.(1), we find E(0)=0

2) For r→∞, the electric field tends to zero as well, because according to eq.(2), the electric field strength decreases with the distance r

3) The maximum electric field occur for $r=r_B$ , i.e. on the surface of the solid ball: in fact, for $r$ the electric field increases with distance, while for $r>r_B$ the field decreases with distance, so the maximum value of the field is for $r=r_B$ .