Question

A $\mathbf{50}\mathbf{}\mathbf{Kg}$satellite circles planet Cruton every 6.0 h.The magnitude of the gravitational force exerted on the satellite by Cruton is 80 N. (a) What is the radius of the orbit? (b) What is the kinetic energy of the satellite? (c) What is the mass of planet Cruton?

Short answer

Answer:

1. The radius of the orbit of the satellite$r=1.9\times {10}^7\text{m}.$
2. The kinetic energy of the satellite$K.E=7.6\times {10}^{8}J.$
3. Mass of the planet Cruton$M=8.6\times {10}^{24}kg$

Step-by-step solution

Identification of given data

The mass of satellite is m = 50 kg

The period of satellite is

$$\begin{gathered} T=6hour \\ =21600s \end{gathered}$$

The magnitude of gravitational force F = 80 N

Significance of centripetal force

A force that causes a body to follow a curved path is known as a centripetal force. It always moves in a direction perpendicular to the body's velocity and in the direction of the instantaneous centre of the path's curvature.

We can use the concept of gravitation and centripetal force.

Formula:

$v=\frac{2\pi r}{T}$ …(i) $F=\frac{m{v}^2}{r}$ …(ii)

Where,F is the centripetal force, v is velocity of object moving in circular direction, m is the mass of object, r is the radius of orbit and T is the time period of object

(a) Determining the orbit of the satellite

According to the equation for equilibrium of forces for the satellite,

Gravitational force = centripetal force.

So,

$F=\frac{m{v}^2}{r}$

Now, using the equation (i) for velocity of satellite, in this equation, we get

$F=\frac{m\left(4{\pi }^{2}r\right)}{T^2}$

Now, rearranging this equation for radius r, we get

$r=\frac{F{T}^2}{4m{\pi }^2}$

Substitute all the values in above equation

role="math" $$\begin{gathered} r=\frac{80 \text{N}\times {\left(\text{21600} \text{s}\right)}^2}{4\times 50 \text{kg}\times {\pi }^2} \\ =1.9\times {10}^{7}m \end{gathered}$$

(b) Determining the kinetic energy of satellite

The equation for kinetic energy is
_{$K.E=\frac{1}{2}m{v}^2$}

Using the equation (ii) for velocity in this equation, we get

$$\begin{gathered} K.E=\frac{1}{2}m\times \frac{Fr}{m} \\ KE=\frac{1}{2}Fr \\ =\frac{1}{2}80 \text{N}\times 1.9\times {10}^7 \text{m} \\ =7.6\times {10}^8\text{J} \end{gathered}$$

Step: 5 (c) Determining the mass of planet

We have the equation of equilibrium of gravitational force and the centripetal force.

$\frac{m{v}^2}{r}=\frac{GMm}{r^2}$

Now, using the equation for velocity and rearranging the equation, we get

$$\begin{gathered} M=\frac{v^{2}r}{G} \\ =\frac{4{\pi }^2{r}^3}{T^{2}G} \\ =\frac{4\times {\pi }^2\times {\left(1.9\times {10}^{7}m\right)}^3}{{\left(\text{21600} \text{s}\right)}^2\times \left(6.67\times {10}^{-11}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)} \\ =8.6\times {10}^{\text{24}}\text{kg} \end{gathered}$$

Mass of the planet Cruton$M=8.6\times {10}^{24}kg$