Question 341: Gravitation

Question

Two satellites of same mass are launched in circular orbits at heights '$$R$$' and '$$2 R$$' above the surface of the earth. The ratio of their kinetic energies is ($$R=$$ radius of the earth)

Accepted Answer

To find the ratio of the kinetic energies of two satellites orbiting at different heights above the Earth, we use the formula for kinetic energy in terms of gravitational force and orbital radius. The formula for the kinetic energy ($$KE$$) of a satellite in orbit around the Earth is given by the expression: $$KE = \frac{GM_Em}{2r}$$ where:
$$KE$$ is the kinetic energy of the satellite,
$$G$$ is the gravitational constant,
$$M_E$$ is the mass of the Earth,
$$m$$ is the mass of the satellite, and
$$r$$ is the distance from the center of the Earth to the satellite (i.e., the Earth's radius plus the height of the satellite above the Earth's surface).
For the first satellite at a height $$R$$ above the surface, the total distance from the center of the Earth is $$r_1 = R + R = 2R$$. Thus, its kinetic energy ($$KE_1$$) would be:
$$KE_1 = \frac{GM_Em}{2(2R)} = \frac{GM_Em}{4R}$$.
For the second satellite at a height $$2R$$ above the surface, the total distance from the center of the Earth is $$r_2 = R + 2R = 3R$$. Thus, its kinetic energy ($$KE_2$$) would be:
$$KE_2 = \frac{GM_Em}{2(3R)} = \frac{GM_Em}{6R}$$.
To find the ratio of their kinetic energies, $$\frac{KE_1}{KE_2}$$, we divide $$KE_1$$ by $$KE_2$$:
$$\frac{KE_1}{KE_2} = \frac{\frac{GM_Em}{4R}}{\frac{GM_Em}{6R}} = \frac{GM_Em}{4R} \cdot \frac{6R}{GM_Em} = \frac{6}{4} = \frac{3}{2}$$.
Therefore, the correct answer is Option B, which shows the ratio of their kinetic energies as $$3:2$$.

Answer

$$1: 3$$

Answer

$$3: 2$$

Answer

$$4: 9$$

Answer

$$9: 4$$

Gravitation

Two satellites of same mass are launched in circular orbits at heights '$R' and '2 R' above the surface of the earth. The ratio of their kinetic energies is (R=$ radius of the earth)

Choose the correct answer: