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 escape velocity of a body depends upon its mass as

Escape speed of a body from Earth's surface is given by: \$v_{min}= \sqrt {2gR}\$As we can see from the above equation, there is no 'mass' term. Implies it can be written as \$m^{0}\$.So, the escape speed of a body is independent of its mass.

Question 1

The value of $g$ at a height of $100km$ from the surface of the Earth is nearly (Radius of the Earth $=$ 6400km) ($g$ on the surface of the Earth $=$ $9.8m/s^{2}$)

Accepted Answer

Acceleration due to gravity changes with the height from Earth's surface as:$g^{'} = g(1-\dfrac {2h}{R})$$\Rightarrow g^{'} = 9.8 (1-\dfrac {2\times 100}{6400})$$\Rightarrow g^{'} = 9.8(1-\dfrac {1}{32})=9.49 m/s^2$

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

A satellite is revolving around the earth in an elliptical orbit. Its speed will be

Accepted Answer

Since the satellite is revolving in elliptical orbit, its distance from the planet changes. So, the speed of the orbit changes.It is slowest at the farthest point and fastest at the nearest point.

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

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 8

The escape velocity of a body depends upon its mass as

Accepted Answer

Escape speed of a body from Earth's surface is given by: $v_{min}= \sqrt {2gR}$As we can see from the above equation, there is no 'mass' term. Implies it can be written as $m^{0}$.So, the escape speed of a body is independent of its mass.

Question 9

The earth retains its atmosphere. This is due to

Accepted Answer

The earth retains its atmosphere. This is due to the escape velocity being greater than the mean speed of the molecules of the atmospheric gases.

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
12841 Views

The value of $g$ at a height of $100 k m$ from the surface of the Earth is nearly (Radius of the Earth $=$ 6400km) ( $g$ on the surface of the Earth $=$ $9.8 m / s^{2}$ )

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 $4 5^{0}$ 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 $= 6400 k m$ )

A satellite is revolving around the earth in an elliptical orbit. Its speed will be

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 $R$ ?

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 escape velocity of a body depends upon its mass as

The earth retains its atmosphere. This is due to

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 12841 Views

The value of g at a height of 100km from the surface of the Earth is nearly (Radius of the Earth = 6400km) (g on the surface of the Earth = 9.8m/s2)

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)

A satellite is revolving around the earth in an elliptical orbit. Its speed will be

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?

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 escape velocity of a body depends upon its mass as

The earth retains its atmosphere. This is due to

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
12841 Views

The value of $g$ at a height of $100 k m$ from the surface of the Earth is nearly (Radius of the Earth $=$ 6400km) ( $g$ on the surface of the Earth $=$ $9.8 m / s^{2}$ )

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 $4 5^{0}$ 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 $R$ ?

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

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