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Chapter 6: Problem 748

The escape velocity for a body projected vertically upwards from the surface of earth is \(11 \mathrm{kms}^{-1}\). If the body is projected at an angle of \(45^{\circ}\) with the vertical, the escape velocity will be $\ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \((11 / \sqrt{2})\) (B) \(11 \sqrt{2}\) (C) 22 (D) 11

Short Answer

Expert verified
The escape velocity for a body projected at an angle of 45° with the vertical is \(\frac{11}{\sqrt{2}}\) km/s. Therefore, the correct answer is (A) \(\frac{11}{\sqrt{2}}\) km/s.

Step by step solution

01

Know the formula of escape velocity

Escape velocity is calculated using the formula: \[v_e = \sqrt{\frac{2GM}{R}}\] where, \(v_e\) is the escape velocity, \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the radius of the Earth.
02

Determine the vertically projected escape velocity

We are given that the escape velocity for a body projected vertically upwards is 11 km/s. Thus, for this case, \(v_e = 11\) km/s.
03

Understand the dependence of escape velocity and projection angle

The escape velocity is directly proportional to the initial velocity along the vertical axis. When the body is projected at an angle of 45°, the vertical component of the initial velocity will be different than when it's projected vertically upwards.
04

Calculate the vertical component of the initial velocity for a 45° projection

When the body is projected at an angle of 45° with the vertical, the vertical component of the initial velocity is given by: \[v_{vertical} = v_e \times \cos{45°}\] As we know that \(\cos{45°} = \frac{1}{\sqrt{2}}\), we can replace it in the equation: \[v_{vertical} = 11 \times \frac{1}{\sqrt{2}}\]
05

Apply the escape velocity formula for the 45° projection

The escape velocity for 45° projection can now be determined using the vertical component of the initial velocity: \[v_e= \sqrt{\frac{2GM}{R}} * v_{vertical}\] Substitute the known values: \[v_e = 11 \times \frac{1}{\sqrt{2}}\] \[v_e = \frac{11}{\sqrt{2}}\, \mathrm{kms}^{-1}\] Thus, the escape velocity for a body projected at an angle of 45° with the vertical is \(\frac{11}{\sqrt{2}}\) km/s. Therefore, the correct answer is (A) \(\frac{11}{\sqrt{2}}\) km/s.

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Most popular questions from this chapter

The mass of a space ship is \(1000 \mathrm{~kg} .\) It is to be launched from earth's surface out into free space the value of \(\mathrm{g}\) and \(\mathrm{R}\) (radius of earth) are \(10 \mathrm{~ms}^{-2}\) and \(6400 \mathrm{~km}\) respectively the required energy for this work will be $=\ldots \ldots \ldots .$ J (A) \(6.4 \times 10^{11}\) (B) \(6.4 \times 10^{8}\) (C) \(6.4 \times 10^{9}\) (D) \(6.4 \times 10^{10}\)

A geo-stationary satellite is orbiting the earth of a height of \(6 \mathrm{R}\) above the surface of earth, \(\mathrm{R}\) being the radius of earth. The time period of another satellite at a height of \(2.5 \mathrm{R}\) from the surface for earth is \(=\ldots \ldots \ldots \ldots\) (A) 6 (B) \(6 \sqrt{2}\) (C) 10 (D) \(6 / \sqrt{2}\)

A body of mass \(\mathrm{m}\) is taken from earth surface to the height \(\mathrm{h}\) equal to radius of earth, the increase in potential energy will be (A) \(\operatorname{mg} R\) (B) \((1 / 2) \mathrm{mgR}\) (C) \(2 \mathrm{mg} \mathrm{R}\) (D) \((1 / 4) \mathrm{mgR}\)

The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density ( \(p\) ), its radius (R) and gravitational constant (G), thus the formula for escape velocity is (A) \(U=\mathrm{R} \sqrt{[}(8 \pi / 3) \mathrm{Gp}]\) (C) \(\mathrm{U}=\sqrt{(2 \mathrm{GMR})}\) (D) \(U=\sqrt{\left[(2 \mathrm{GMR}) / \mathrm{R}^{2}\right]}\)

At what distance from the center of earth, the value of acceleration due to gravity \(g\) will be half that of the surfaces \((\mathrm{R}=\) Radius of earth \()\) (A) \(2 \mathrm{R}\) (B) \(\mathrm{R}\) (C) \(1.414 \mathrm{R}\) (D) \(0.414 \mathrm{R}\)
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