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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

A particle of mass \(10 \mathrm{~g}\) is kept on the surface of a uniform sphere of mass \(100 \mathrm{~kg}\) and radius \(10 \mathrm{~cm}\). Find the work to be done against the gravitational force between them to take the particle is away from the sphere \(\left(\mathrm{G}=6.67 \times 10^{-11} \mathrm{SI}\right.\) unit \()\) (A) \(6.67 \times 10^{-9} \mathrm{~J}\) (B) \(6.67 \times 10^{-10} \mathrm{~J}\) (C) \(13.34 \times 10^{-10} \mathrm{~J}\) (D) \(3.33 \times 10^{-10} \mathrm{~J}\)

A planet moving along an elliptical orbit is closest to the sun at a distance \(\mathrm{r}_{1}\) and farthest away at a distance of \(\mathrm{r}_{2}\). If \(\mathrm{v}_{1}\) and \(\mathrm{v}_{2}\) are the liner velocities at these points respectively, then the ratio \(\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\) is \(\ldots \ldots \ldots \ldots\) (A) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\) (B) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)^{2}\) (C) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)\) (D) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)^{2}\)

Escape velocity on the surface of earth is \(11.2 \mathrm{kms}^{-1}\) Escape velocity from a planet whose masses the same as that of earth and radius $1 / 4\( that of earth is \)=\ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \(2.8\) (B) \(15.6\) (C) \(22.4\) (D) \(44.8\)

The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is \(\mathrm{V}_{\mathrm{e}}\) on the earth (A) \(\mathrm{V}_{\mathrm{e}}\) (B) \(2 \mathrm{~V}_{\mathrm{e}}\) (C) \(4 \mathrm{~V}_{\mathrm{e}}\) (D) \(\mathrm{V}_{\mathrm{e}} / 2\)

Density of the earth is doubled keeping its radius constant then acceleration, due to gravity will be \(-m s^{-2}\) \(\left(\mathrm{g}=9.8 \mathrm{~ms}^{2}\right)\) (A) \(19.6\) (B) \(9.8\) (C) \(4.9\) (D) \(2.45\)
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