Gravitation Test

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Gravitation Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of escape velocity of a tennis ball to that of a basket ball on the surface of the earth is

A
1:1

B
1:2

C
2:1

D
4:1

Solution

Equating KE and PE we get the expression for escape velocity as $$\sqrt{2GM/r}$$, where M = mass of earth , r = radius of earth and G = gravitational constant. ince escape velocity is independent of mass of ball, the ratio of escape velocity of a tennis ball to that of a basket ball on the surface of the earth is 1:1.
Question 2
1 / -0

A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth

A
the acceleration of S is always directed towards the centre of the earth

B
the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant

C
the total mechanical energy of S varies periodically with time

D
the linear momentum of S remains constant in magnitude

Solution

In case of orbital motion, only angular momentum is conserved, i.e. mvr is constant. Thus v,r changes accordingly. Therefore, linear momentum mv is not constant in elliptical orbit. So option D is wrong
The total mechanical energy is given by $$ \dfrac{-GMm}{2r} $$ which is independent of time. So, option C is wrong
Both, direction and magnitude of angular momentum does not change. i.e. $$ r \times mv$$ is constant. So, B is also wrong.
The gravitational force always towards the centre of the earth and thus the acceleration is always directed to the centre of earth. So, A is right.
Question 3
1 / -0

The distance from the surface of the earth at which above and below the surface acceleration due to gravity is same will be

A
$$\displaystyle h = \frac{\sqrt{5}- 1}{2} R$$

B
$$\displaystyle h = \frac{\sqrt{3} - 1}{2}R$$

C
$$\displaystyle h = \frac{\sqrt{3} - 1}{3} R$$

D
None of these

Solution

Let the distance be $$d$$.
For the condition:
$$g\left(1-\dfrac{ h }{ R }\right) =g\left(\dfrac { R }{ R+h } \right)^2 $$
$$\left(1-\dfrac { h }{ R }\right) \left(1+\dfrac { h }{ R }\right)^{ 2 }=1$$
Solving this we get:
$$h=\dfrac { \sqrt { 5 } -1 }{ 2 } R$$
Question 4
1 / -0

A satellite in earth orbit experiences a small drag force as it enters the earth's atmosphere. Two students were asked consequence of this
Student-A : The satellite would slow down as, it spirals towards earth due to work of frictional force.
Student-B : The satellite speed up due to earths gravitational pull as it spirals towards earth.

A
A is correct, B is wrong

B
B is correct, A is wrong

C
both are correct

D
both are wrong

Solution

Let, $$U + K = c$$ where c is some constant. As the body spirals down, $$ U \downarrow, K \uparrow$$. Thus, speed of the satellite increases and the influence of air friction is negligable.
Question 5
1 / -0

Satellites A and B are orbiting around the earth in orbits of ratio $$R$$ and $$4R$$ respectively. The ratio of their areal velocities is

A
$$1 : 2$$

B
$$1 : 4$$

C
$$1 : 8$$

D
$$1 : 16$$

Solution

Using Kepler's second law we have:
$$Areal \;velocity =\dfrac{Area swept}{time}=\dfrac{\dfrac{1}{2}r(rd\theta)}{dt}=\dfrac{1}{2}r^2\omega=\dfrac{1}{2}rv=constant$$

Thus we get, $$\dfrac{1}{2}\sqrt{\dfrac{GM}{r}}r$$ (from $$\dfrac{mv^2}{r}=\dfrac{GMm}{r^2}$$)
Thus we get
Areal velocity $$\propto \sqrt{r}$$
Thus for the planets with radius in ratio of $$1:4$$ we get radii in the ratio of $$1:2$$
Question 6
1 / -0

A planet of radius $$R = 1/10 \times $$ (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth $$R/5$$ on it and lower a wire of the same length and of linear mass density $$10^{-3} kgm^{-3}$$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth $$= 6 \times 10^6 m$$ and the acceleration due to gravity on Earth is $$10 ms^{-2}$$)

A
$$96 N$$

B
$$108 N$$

C
$$120 N$$

D
$$150 N$$

Solution

$$R_e$$ is the radius of the earth
$$g_e=\dfrac{GM_e}{R_e^2}=G\dfrac{4}{3}\pi R_e\rho$$
Now g at the mentioned planet is given as
$$g_p=\dfrac{GM}{r^2}$$, here $$m=\dfrac{4}{3}\pi r^3\rho$$
Thus $$g_p=\dfrac{4}{3}r\rho$$
Thus we get $$F=\displaystyle\int^R_{4R/5}\lambda g_p dr = 108 N$$
Question 7
1 / -0

Two planets $$A$$ and $$B$$ have the same material density. If the radius of $$A$$ is twice that of B, then the ratio of the escape velocity $$\displaystyle \dfrac{V_A}{V_B}$$ is

A
$$2$$

B
$$\sqrt{2}$$

C
$$\dfrac{1}{ \sqrt{2}}$$

D
$$\dfrac{1}{2}$$

Solution

Let the density be d for both the planets. Given that $$R_A = 2R_B$$
Now, mass of $$A,\ M_A = \dfrac{4d\pi {R_A}^3}{3} = \dfrac{32d\pi {R_B}^3}{3}$$
similarly, $$M_B = \dfrac{4d\pi {R_B}^3}{3}$$
Escape velocity for a planet is given by $$V =\sqrt{\dfrac{2GM}{R}}$$

So, $$V_A = \sqrt{\dfrac{2GM_A}{3R_A}} = \sqrt{\dfrac{64Gd\pi {R_B}^3}{6R_B}} = \sqrt{\dfrac{32Gd\pi {R_B}^2}{3}}$$

Similarly, $$V_B = \sqrt{\dfrac{8Gd\pi {R_B}^2}{3}}$$

Taking the ratio, $$\dfrac{V_A}{V_B} = \sqrt{\dfrac{32Gd\pi {R_B}^2}{3}} \times \sqrt{\dfrac{3}{8Gd\pi {R_B}^2}} =2$$
Question 8
1 / -0

A planet revolves about the sun in elliptical orbit. The arial velocity $$\displaystyle \left ( \frac{dA}{dt} \right )$$ of the planet is $$4.0 \times 10^{16} m^2/s$$. The least distance between planet and the sun is $$2 \times 10^{12}$$m. Then the maximum speed of the planet in km/s is

A
10

B
20

C
40

D
none of these

Solution

We have: $$\dfrac{dA}{dt}=\dfrac{1}{2}r^2\dfrac{d\theta}{dt}$$

$$4\times 10^{16}=\dfrac{1}{2}(2\times 10^{12})^2\dfrac{d\theta}{dt}$$

$$\omega=\dfrac{d\theta}{dt}=2\times 10^{-8}$$

Using, $$v=r\omega$$
We get: $$v=2\times 10^{12}\times 2\times 10^{-8}=40\ km/s $$
Question 9
1 / -0

A (nonrotating) star collapses onto itself from an initial radius R$$_i$$ with its mass remaining unchanged. Which curve in figure best gives the gravitational acceleration a$$_g$$ on the surface of the star as a function of the radius of the star during the collapse?

A
a

B
b

C
c

D
d

Solution

$$g\propto \dfrac{1}{R^2}$$
Curve b represents this variation.
Question 10
1 / -0

Select the correct choice(s)

A
The gravitational field inside a spherical cavity, within a spherical planet must be non zero and uniform.

B
When a body is projected horizontally at an appreciable large height above the earth, with a velocity less than for a circular orbit, it will fall to the earth along a parabolic path.

C
A body of zero total mechanical energy placed in a gravitational field will escape the field

D
Earth's satellite must be in equatorial plane.

Solution

Gravitational field is conservative field, total mechanical energy is conserved.
If a body is in gravitational field with $$ U + K = 0 \rightarrow K = -U \rightarrow (1)$$ i.e. Kinetic energy is non zero.So, it will move away from the field till its kinetic energy is also zero.
If $$K = 0 \rightarrow$$ U = 0 from (1). But, U = 0 is at $$\infty$$ and thus the body escapes the field.

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Gravitation Test - 29

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Gravitation Test - 29

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Question 1

1:1

1:2

2:1

4:1

Solution

Question 2

the acceleration of S is always directed towards the centre of the earth

the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant

the total mechanical energy of S varies periodically with time

the linear momentum of S remains constant in magnitude

Solution

Question 3

$$\displaystyle h = \frac{\sqrt{5}- 1}{2} R$$

$$\displaystyle h = \frac{\sqrt{3} - 1}{2}R$$

$$\displaystyle h = \frac{\sqrt{3} - 1}{3} R$$

None of these

Solution

Question 4

A is correct, B is wrong

B is correct, A is wrong

both are correct

both are wrong

Solution

Question 5

$$1 : 2$$

$$1 : 4$$

$$1 : 8$$

$$1 : 16$$

Solution

Question 6

$$96 N$$

$$108 N$$

$$120 N$$

$$150 N$$

Solution

Question 7

$$2$$

$$\sqrt{2}$$

$$\dfrac{1}{ \sqrt{2}}$$

$$\dfrac{1}{2}$$

Solution

Question 8

10

20

40

none of these

Solution

Question 9

a

b

c

d

Solution

Question 10

The gravitational field inside a spherical cavity, within a spherical planet must be non zero and uniform.

When a body is projected horizontally at an appreciable large height above the earth, with a velocity less than for a circular orbit, it will fall to the earth along a parabolic path.

A body of zero total mechanical energy placed in a gravitational field will escape the field

Earth's satellite must be in equatorial plane.

Solution

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