Question 359: Gravitation

Question

The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is '$$V_e$$', then the escape velocity from the planet is

Accepted Answer

To find the escape velocity from the planet, we need to use the formula for escape velocity. The escape velocity $$V$$ from a celestial body is given by:
$$V = \sqrt{\frac{2GM}{R}}$$
where:
$$G$$ is the gravitational constant,
$$M$$ is the mass of the celestial body, and
$$R$$ is the radius of the celestial body.
For Earth, the escape velocity is:
$$V_e = \sqrt{\frac{2GM_e}{R_e}}$$
where $$M_e$$ is the mass of the Earth and $$R_e$$ is the radius of the Earth.
Now, for the planet in question, the mass of the planet $$M_p$$ is 6 times that of the Earth, i.e.,
$$M_p = 6M_e$$
The radius of the planet $$R_p$$ is twice that of the Earth, i.e.,
$$R_p = 2R_e$$
The escape velocity $$V_p$$ from the planet is then given by:
$$V_p = \sqrt{\frac{2GM_p}{R_p}}$$
Substituting the values of $$M_p$$ and $$R_p$$ in the escape velocity formula, we get:
$$V_p = \sqrt{\frac{2G(6M_e)}{2R_e}}$$
Simplifying the expression inside the square root:
$$V_p = \sqrt{\frac{12GM_e}{2R_e}}$$
$$V_p = \sqrt{6 \frac{GM_e}{R_e}}$$
Since we know that:
$$V_e = \sqrt{\frac{2GM_e}{R_e}}$$
we can substitute $$V_e^2$$ in place of $$\frac{2GM_e}{R_e}$$:
$$V_e^2 = \frac{2GM_e}{R_e}$$
Therefore:
$$V_p = \sqrt{3 \cdot V_e^2}$$
$$V_p = \sqrt{3} \cdot V_e$$
Thus, the escape velocity from the planet is $$\sqrt{3} \cdot V_e$$.
The correct answer is Option A: $$\sqrt{3} \mathrm{~V}_{\mathrm{e}}$$.

Answer

$$\sqrt{3} \mathrm{~V}_{\mathrm{e}}$$

Answer

$$\sqrt{2} \mathrm{~V}_{\mathrm{e}}$$

Answer

$$\mathrm{V}_{\mathrm{e}}$$

Answer

$$\sqrt{5} \mathrm{~V}_{\mathrm{e}}$$

Gravitation

The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is '$V_e$', then the escape velocity from the planet is

Choose the correct answer: