Q: The escape velocity for a body projected vertically upwards from the surface of earth is \$11\$ km/s. If the body is projected at an angle of \$45^{0}\$ with the vertical, the escape velocity will be

Escape speed of a body from Earth's surface is given by: \$v_{min}= \sqrt {2gR}\$This expression is obtained by conservation of energy and doesn't involve in which direction the body is thrown/projected.So, irrespective of the angle of projection, escape speed of the body from Earth's surface remains constant i.e. \$\approx 11\$ km/s

Q: If \$g\$ on the surface of the Earth is \$9.8\ ms^{-2}\$, its value at a height of \$6400 km\$ is: (Radius of the Earth \$= 6400km\$)

The expression for \$g\$ is given by:\$g_R =\dfrac {GM}{R^2}\$The value of \$g\$ changes according to\$g_h=\dfrac {GM}{(R+h)^2}\$Therefore, \$\displaystyle \dfrac{g_{h}}{g_R} = \dfrac{R^{2}}{(R+h)^{2}}\$Here, \$h = R\$So, \$\dfrac {g_h}{g_R}=(\dfrac {R}{R+h})^2\$\$\Rightarrow \dfrac {g_h}{g_R}=(\dfrac {R}{R+R})^2\$\$\Rightarrow g_h=g_R\times\dfrac {1}{4} = 9.8\times 0.25=2.45\ m/s^2\$

Q: At what height, the value of '\$ g \$' is half that on the surface of the earth of radius \$R\$?

We have, \$g=\dfrac {GM}{r^2}\$; \$r\$ is the distance from the centre of Earth.\$\Rightarrow g\propto \dfrac {1}{r^2}\$ above the surface of the earth.\$\Rightarrow \dfrac {g_s}{g_h}=\dfrac {r_h^2}{r_s^2}\$Since we want the value of \$g_s\$ to be half, we replace \$g_h\$ by \$g_s/2\$\$\Rightarrow \dfrac {g_s}{g_s/2}=[\dfrac {R+h}{R}]^2\$\$\Rightarrow 2=[\dfrac {R+h}{R}]^2\$\$\Rightarrow \dfrac {R+h}{R}=\sqrt 2\$\$\Rightarrow \dfrac {h}{R}=1.414-1=0.414\$\$\Rightarrow h=0.414R\$

Q: The ratio of escape velocities of two planets if \$g\$ values on the two planets are \$9.9\$ m/s\$^{2}\$ and \$3.3\$ m/s\$^{2}\$ and their radii are \$6400km\$ and \$3400km\$ respectively is

Escape speed, \$v_e=\sqrt {2gR}\$\$\Rightarrow \dfrac {v_2}{v_1}=\sqrt {\dfrac {g_2R_2}{g_1R_1}}=\sqrt {\dfrac {9.9\times6400}{3.3\times3400}}=2.37\$\$\Rightarrow v_2:v_1=2.37:1\$

Question 1

The escape velocity of a sphere of mass $m$ is given by

Accepted Answer

For the body to escape Earth's gravitation and reach infinity, the initial kinetic energy has to be just higher than the Gravitational potential energy of the body on the surface.Total energy of the body just after projecting is:$E=\dfrac {1}{2}mv^2 - \dfrac {GMm}{R_e}$Now this total energy has to be just greater than zero, so the minimum speed is achieved when kinetic energy is equal to gravitational potential energy.$\Rightarrow \dfrac {1}{2}mv^2=\dfrac {GMm}{R_e}$$\Rightarrow \dfrac {1}{2}v^2=\dfrac {GM}{R_e}$$\Rightarrow v=\sqrt {\dfrac {2GM}{R_e}}$

Question 2

The escape velocity for a body projected vertically upwards from the surface of earth is $11$ km/s. If the body is projected at an angle of $45^{0}$ with the vertical, the escape velocity will be

Accepted Answer

Escape speed of a body from Earth's surface is given by: $v_{min}= \sqrt {2gR}$This expression is obtained by conservation of energy and doesn't involve in which direction the body is thrown/projected.So, irrespective of the angle of projection, escape speed of the body from Earth's surface remains constant i.e. $\approx 11$ km/s

Question 3

If $g$ on the surface of the Earth is $9.8\  ms^{-2}$, its value at a height of $6400 km$ is: (Radius of the Earth $= 6400km$)

Accepted Answer

The expression for $g$ is given by:$g_R =\dfrac {GM}{R^2}$The value of $g$ changes according to$g_h=\dfrac {GM}{(R+h)^2}$Therefore, $\displaystyle \dfrac{g_{h}}{g_R} = \dfrac{R^{2}}{(R+h)^{2}}$Here, $h = R$So, $\dfrac {g_h}{g_R}=(\dfrac {R}{R+h})^2$$\Rightarrow \dfrac {g_h}{g_R}=(\dfrac {R}{R+R})^2$$\Rightarrow g_h=g_R\times\dfrac {1}{4} = 9.8\times 0.25=2.45\ m/s^2$

Question 4

Which of the following quantities remain constant in a planetary motion, when seen from the surface of the sun?

Accepted Answer

When viewed from the inertial frame of the sun, the planets seem to execute elliptical orbital motion with changing ($K.E$) (Kepler's first law) but constant angular momentum (due to the force being central).

Question 5

The gravitational field is a conservative field. The work done in this field by moving an object from one point to another

Accepted Answer

A conservative  force is a type of force wherein there is no net work done during its motion in any closed loop.When we throw a ball up there is negative work done as it moves against the force of gravity and during the fall the gravity is in the positive direction. The resultant work done is zero and the force of gravity is path independent.Thus, gravitational field is a conservative field and  work done depends on the end points only.

Question 6

At what height, the value of '$ g $' is half that on the surface of the earth of radius $R$?

Accepted Answer

We have, $g=\dfrac {GM}{r^2}$; $r$ is the distance from the centre of Earth.$\Rightarrow g\propto \dfrac {1}{r^2}$ above the surface of the earth.$\Rightarrow \dfrac {g_s}{g_h}=\dfrac {r_h^2}{r_s^2}$Since we want the value of $g_s$ to be half, we replace $g_h$ by $g_s/2$$\Rightarrow \dfrac {g_s}{g_s/2}=[\dfrac {R+h}{R}]^2$$\Rightarrow 2=[\dfrac {R+h}{R}]^2$$\Rightarrow \dfrac {R+h}{R}=\sqrt 2$$\Rightarrow \dfrac {h}{R}=1.414-1=0.414$$\Rightarrow h=0.414R$

Question 7

The ratio of escape velocities of two planets if $g$ values on the two planets are $9.9$ m/s$^{2}$ and $3.3$ m/s$^{2}$ and their radii are $6400km$ and $3400km$ respectively is

Accepted Answer

Escape speed, $v_e=\sqrt {2gR}$$\Rightarrow \dfrac {v_2}{v_1}=\sqrt {\dfrac {g_2R_2}{g_1R_1}}=\sqrt {\dfrac {9.9\times6400}{3.3\times3400}}=2.37$$\Rightarrow v_2:v_1=2.37:1$

Question 8

A gravitational field is present in a region. A point mass is shifted from $A$ to $B$ , along different paths shown in the figure. If $W_{1}$ , $W_{2}$ and $W_{3}$ represent the work done by gravitational force for respective paths, then

Accepted Answer

Work done is independent of the distance traveled or the path chosen. It depends only on the displacement. Since for all the paths given, the displacement is same, the work done by gravitational field in each case is equal.

Question 9

The kinetic energy needed to project a body of mass $m$ from earth's surface $($ radius $R$  $)$ to infinity is

Accepted Answer

For the body to escape Earth's gravitation and reach infinity, the initial kinetic energy has to be just higher than the Gravitational potential energy of the body on the surface.Total energy of the body just after projecting is:$E=\dfrac {1}{2}mv^2 - \dfrac {GMm}{R}= \dfrac {1}{2}mv^2 - mgR$To escape Earth's gravitation, this energy has to be positive:$\Rightarrow \dfrac {1}{2}mv^2 - mgR \geq 0\Rightarrow K.E \geq mgR$So, the minimum Kinetic energy required is $mgR$.

Question 10

If $g$ on the surface of the Earth is $9.8$ $ms^{-2}$, then it's value at a depth of $3200$ $km$ (Radius of the earth $ =  6400$ $km$) is

Accepted Answer

The value of gravity changes as we move away from or towards the centre of the Earth.This is given by: $g_{R} =\dfrac{GM}{ R^{2} }$, where $M$ is the mass of a planet of radius $R$So, $M=\dfrac{4}{3}\pi R^{3}\rho$ ; substituting in above equation $\Rightarrow g_R = G \dfrac{4}{3} \pi \rho \times R $Since we want the value of $g$ at depth($d$) from the Earth's surface, we replace $R$ by $(R-d)$$\Rightarrow g_d = G \dfrac{4}{3} \pi \rho \times (R-d) $$\Rightarrow \dfrac{g_{R}}{ g_{d} } = \dfrac{R}{R-d} $$\Rightarrow  \dfrac{g_{d}}{ g_{R} } =(1- \dfrac{d}{R})$The given depth in the problem is $d = 3200$  km, substituting we get,$ \dfrac{g_{d}}{ g_{R} } =(1- \dfrac{3200}{6400})$$ {g_{d}}=9.8\times(1- \dfrac{3200}{6400})$$ {g_{d}}=9.8/2$$ {g_{d}}=4.9 m s^{-2} $

Gravitation

Physics
3390 Views

The escape velocity of a sphere of mass $m$ is given by

The escape velocity for a body projected vertically upwards from the surface of earth is $11$ km/s. If the body is projected at an angle of $4 5^{0}$ with the vertical, the escape velocity will be

If $g$ on the surface of the Earth is $9.8 m s^{- 2}$ , its value at a height of $6400 k m$ is: (Radius of the Earth $= 6400 k m$ )

Which of the following quantities remain constant in a planetary motion, when seen from the surface of the sun?

The gravitational field is a conservative field. The work done in this field by moving an object from one point to another

At what height, the value of ' $g$ ' is half that on the surface of the earth of radius $R$ ?

The ratio of escape velocities of two planets if $g$ values on the two planets are $9.9$ m/s $^{2}$ and $3.3$ m/s $^{2}$ and their radii are $6400 k m$ and $3400 k m$ respectively is

A gravitational field is present in a region. A point mass is shifted from $A$ to $B$ , along different paths shown in the figure. If $W_{1}$ , $W_{2}$ and $W_{3}$ represent the work done by gravitational force for respective paths, then

The kinetic energy needed to project a body of mass $m$ from earth's surface $($ radius $R$ $)$ to infinity is

If $g$ on the surface of the Earth is $9.8$ $m s^{- 2}$ , then it's value at a depth of $3200$ $k m$ (Radius of the earth $= 6400$ $k m$ ) is

Gravitation

Physics 3390 Views

The escape velocity of a sphere of mass m is given by

The escape velocity for a body projected vertically upwards from the surface of earth is 11 km/s. If the body is projected at an angle of 450 with the vertical, the escape velocity will be

If g on the surface of the Earth is 9.8 ms−2, its value at a height of 6400km is: (Radius of the Earth =6400km)

Which of the following quantities remain constant in a planetary motion, when seen from the surface of the sun?

The gravitational field is a conservative field. The work done in this field by moving an object from one point to another

At what height, the value of 'g' is half that on the surface of the earth of radius R?

The ratio of escape velocities of two planets if g values on the two planets are 9.9 m/s2 and 3.3 m/s2 and their radii are 6400km and 3400km respectively is

A gravitational field is present in a region. A point mass is shifted from A to B, along different paths shown in the figure. If W1​ , W2​ and W3​ represent the work done by gravitational force for respective paths, then

The kinetic energy needed to project a body of mass m from earth's surface ( radius R ) to infinity is

If g on the surface of the Earth is 9.8 ms−2, then it's value at a depth of 3200 km (Radius of the earth =6400 km) is

Physics
3390 Views

The escape velocity of a sphere of mass $m$ is given by

The escape velocity for a body projected vertically upwards from the surface of earth is $11$ km/s. If the body is projected at an angle of $4 5^{0}$ with the vertical, the escape velocity will be

If $g$ on the surface of the Earth is $9.8 m s^{- 2}$ , its value at a height of $6400 k m$ is: (Radius of the Earth $= 6400 k m$ )

At what height, the value of ' $g$ ' is half that on the surface of the earth of radius $R$ ?

The ratio of escape velocities of two planets if $g$ values on the two planets are $9.9$ m/s $^{2}$ and $3.3$ m/s $^{2}$ and their radii are $6400 k m$ and $3400 k m$ respectively is

A gravitational field is present in a region. A point mass is shifted from $A$ to $B$ , along different paths shown in the figure. If $W_{1}$ , $W_{2}$ and $W_{3}$ represent the work done by gravitational force for respective paths, then

The kinetic energy needed to project a body of mass $m$ from earth's surface $($ radius $R$ $)$ to infinity is

If $g$ on the surface of the Earth is $9.8$ $m s^{- 2}$ , then it's value at a depth of $3200$ $k m$ (Radius of the earth $= 6400$ $k m$ ) is