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Chapter 6: Problem 743

The escape velocity of a body on the surface of the earth is $11.2 \mathrm{~km} / \mathrm{sec}$. If the mass of the earth is increases to twice its present value and the radius of the earth becomes half, the escape velocity becomes \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\) \((\Delta) 56\)

Short Answer

Expert verified
The new escape velocity when the Earth's mass is doubled and the radius becomes half is \(44.8 \mathrm{~km/sec}\).

Step by step solution

01

Find the initial escape velocity

We are given that the escape velocity on the surface of the Earth is currently 11.2 km/sec. Convert this to meters per second for calculations: \[v_e = 11.2 \times 1000 \mathrm{~m/sec}\]
02

Calculate the gravitational constant times the Earth's mass

Using the escape velocity formula and the given Earth's radius \(R\) as 6400km or 6400,000m, we can solve for the product of the mass and the gravitational constant as follows: \[\frac{GM}{R} = \frac{v_e^2}{2\RE\] \[GM = (v_e^2)(\frac{R}{2})\]
03

Write the equation for the new escape velocity

Let's denote the new mass as \(M'\), which is twice the current mass, and the new radius as \(R'\), which is half the current radius. Then, the escape velocity formula can be written as: \[v'_e = \sqrt{\frac{2 GM'}{R'}}\]
04

Solve for the new escape velocity

Substituting the values of \(M'\) and \(R'\) into the new escape velocity formula, we get: \[v'_e = \sqrt{\frac{2 G(2M)}{\frac{R}{2}}}\] \[v'_e = \sqrt{\frac{4 GM}{\frac{R}{2}}}\] \[v'_e = \sqrt{\frac{4 (v_e^2)(\frac{R}{2})}{\frac{R}{2}}}\] As we plug in the initial escape velocity value from Step 1, and cancel terms as needed: \[v'_e = \sqrt{\frac{4 (11.2 \times 1000)^2}{1}}\] \[v'_e = 4 \times 11.2 \times 1000 \mathrm{~m/sec}\] \[v'_e = 44.8 \times 1000 \mathrm{~m/sec}\]
05

Convert the new escape velocity to km/sec

Finally, convert the new escape velocity back to km/sec: \[v'_e = 44.8 \mathrm{~km/sec}\] The new escape velocity is 44.8 km/sec.

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Most popular questions from this chapter

The period of a satellite in circular orbit around a planet is independent of (A) the mass of the planet (B) the radius of the planet (C) mass of the satellite (D) all the three parameters (A), (B) and (C)

A satellite with K.E. \(E_{\mathrm{k}}\) is revolving round the earth in a circular orbit. How much more K.E. should be given to it so that it may just escape into outer space ? (A) \(\mathrm{E}_{\mathrm{k}}\) (B) \(2 \mathrm{E}_{\mathrm{k}}\) (C) \((1 / 2) \mathrm{E}_{\mathrm{k}}\) (D) \(3 \mathrm{E}_{\mathrm{k}}\)

Which one of following statements regarding artificial satellite of earth is incorrect (A) The orbital velocity depends on the mass of the satellite (B) A minimum velocity of \(8 \mathrm{kms}^{-1}\) is required by a satellite to orbit quite close to the earth. (C) The period of revolution is large if the radius of its orbit is large (D) The height of geostationary satellite is about \(36000 \mathrm{~km}\) from earth

If mass of a body is \(\mathrm{M}\) on the earth surface, than the mass of the same body on the moon surface is (A) \(\mathrm{M} / 6\) (B) 56 (C) \(\mathrm{M}\) (D) None of these

As astronaut orbiting the earth in a circular orbit \(120 \mathrm{~km}\) above the surface of earth, gently drops a spoon out of space-ship. The spoon will (A) Fall vertically down to the earth (B) move towards the moon (C) Will move along with space-ship (D) Will move in an irregular wav then fall down to earth
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