Question

The escape velocity for earth is v. A planet having 9 times mass that of earth and radius, 16 times that of earth, has the escape velocity of:

• A. \(\frac{v}{3}\)
• B. \(\frac{2v}{3}\)
• C. \(\frac{3v}{4}\)
• D. \(\frac{9v}{4}\)

Answer 1

The Correct Option is C

The escape velocity (\(v_e\)) of a celestial body is the minimum velocity needed for an object to escape the gravitational influence of that body. The formula for escape velocity is given by:

\(v_e = \sqrt{\frac{2GM}{R}}\) 

where:

• \(G\) is the universal gravitational constant
• \(M\) is the mass of the celestial body
• \(R\) is the radius of the celestial body

For Earth, this escape velocity is given as \(v\), i.e.,

\(v = \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}}\)

Now, consider a planet with mass \(M_{\text{planet}}\) and radius \(R_{\text{planet}}\). According to the question:

• \(M_{\text{planet}} = 9M_{\text{Earth}}\)
• \(R_{\text{planet}} = 16R_{\text{Earth}}\)

The escape velocity for the planet would be:

\(v_{\text{planet}} = \sqrt{\frac{2G(9M_{\text{Earth}})}{16R_{\text{Earth}}}}\)

Simplify the equation:

\(v_{\text{planet}} = \sqrt{\frac{9}{16} \cdot \frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}}\)

Since \(v = \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}}\), substitute for \(v\):

\(v_{\text{planet}} = \sqrt{\frac{9}{16}} \cdot v\)

\(v_{\text{planet}} = \frac{3}{4} \cdot v\)

Thus, the escape velocity of the planet is \(\frac{3v}{4}\).

Hence, the correct answer is \(\frac{3v}{4}\).