Q: The ratio of escape velocity at earth\$(v_e)\$ to the escape velocity at a planet\$(v_p)\$ whose radius and mean density are twice as that of earth is:

We know that: \$v_e=\sqrt{\dfrac{2GM}{R}}\$Given that: \$R'=2R\$ and \$\rho'=2\rho\$\$\Rightarrow M=\dfrac{4}{3}\pi R^3\times \rho\$\$\Rightarrow M'=\dfrac{4}{3}\pi(2R)^3\times 2\rho=16M\$\$\Rightarrow v_p=\sqrt{\dfrac{2G(16M)}{2R}}=\sqrt{\dfrac{16GM}{R}}\$\$\Rightarrow \dfrac{v_e}{v_p}=\sqrt{\dfrac{1}{8}}=\dfrac{1}{2\sqrt{2}}\$

Q: The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

Acceteration due to gravity at height h, \$g_n=g_0 (1-\dfrac{2h}{R})\$ \$ h=1km\$Acceleration due to gravity at depth d,\$d_d = g_0\left(1-\dfrac{d}{R}\right)\$\$g_h=g_d\$\$g_0(1-\dfrac{2h}{R})=g_0(1-\dfrac{d}{R})\$\$\rightarrow d=2h\$\$= 2\times 1 km\$\$d=2km\$

Q: Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by which of the following options ?

Acceleration due to gravity, \$g = \dfrac{GMr}{R^3}\$ for \$0\leq r\leq R\$\$\implies\$ \$g \propto r\$ for \$0\leq r\leq R\$Acceleration due to gravity, \$g = \dfrac{GM}{r^2}\$ for \$ r\geq R\$\$\implies\$ \$g \propto \dfrac{1}{r^2} \$ for \$ r\geq R\$Hence option C is correct.

Q: If \${v}_{e}\$ is escape velocity and \${v}_{o}\$ is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by :

\${v}_{escape}=\sqrt {\cfrac{2GM}{R}}\$\${v}_{orbital}=\sqrt {\cfrac{GM}{R}}\$\${v}_{escape}=\sqrt {2}{v}_{o}\$

Q: A particle of mass \$m\$ is thrown upwards from the surface of the earth, with a velocity \$u\$. The mass and the radius of the earth are, respectively, \$M\$ and \$R\$. \$G\$ is gravitational constant and \$g\$ is acceleration due to gravity on the surface of the earth. The minimum value of \$u\$ so that the particle does not return back to earth is

\$GM = g{R}^{2}\$\${V}_{e} = \sqrt{2gR} = \sqrt{2\displaystyle\frac{GM}{{R}^{2}}}R = \sqrt{\displaystyle\frac{2GM}{R}}\$.

Question 1

A body weighs $72 N$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Accepted Answer

Hint: Use formula of acceleration due to gravity at height hExplanation:Weight of the body on earth is given 72 N which is product of mass and acceleration due to gravity.So, $mg = 72$.Acceleration due to gravity at height h is given by the formula, $g&#8217; = g \dfrac{g{R_E}^2}{{{(R}_E+\dfrac{R_E}{2})}^2}$, where $R_E$ is radius of earth. $g&#8217; = \dfrac{4}{9} g$So, weight on height h will be, $W&#8217; = mg&#8217; = 72 * \dfrac{4}{9} = 32 N$$Answer:$Hence, option C is the correct answer.

Question 2

A body of mass '$m$' taken from the earth's surface to the height equal to twice the radius ($R$) of the earth. The change in potential energy of body will be

Accepted Answer

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Question 3

A body weighs 200 N on the surface of the earth. How much it weigh half way down to the centre of the earth ?

Accepted Answer

Acceleration due to gravity at a depth d from surface of earth $g' = g \left(1 - \dfrac{d}{R} \right) $ ....(1)Where g = acceleration due to gravity at earth's surface Multiplying by mass 'm' on both sides of (1)$\implies mg' = mg \left(1 - \dfrac{d}{R} \right) \, \,\, \, Here, \left(d = \dfrac{R}{2} \right)$$= 200 \left(1 - \dfrac{R}{2 R} \right) = \dfrac{200}{2} = 100 N$

Question 4

The ratio of escape velocity at earth$(v_e)$ to the escape velocity at a planet$(v_p)$ whose radius and mean density are twice as that of earth is:

Accepted Answer

We know that: $v_e=\sqrt{\dfrac{2GM}{R}}$Given that: $R'=2R$ and $\rho'=2\rho$$\Rightarrow M=\dfrac{4}{3}\pi R^3\times \rho$$\Rightarrow M'=\dfrac{4}{3}\pi(2R)^3\times 2\rho=16M$$\Rightarrow v_p=\sqrt{\dfrac{2G(16M)}{2R}}=\sqrt{\dfrac{16GM}{R}}$$\Rightarrow \dfrac{v_e}{v_p}=\sqrt{\dfrac{1}{8}}=\dfrac{1}{2\sqrt{2}}$

Question 5

The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

Accepted Answer

Acceteration due to gravity at height h, $g_n=g_0 (1-\dfrac{2h}{R})$  $ h=1km$Acceleration due to gravity at depth d,$d_d = g_0\left(1-\dfrac{d}{R}\right)$$g_h=g_d$$g_0(1-\dfrac{2h}{R})=g_0(1-\dfrac{d}{R})$$\rightarrow d=2h$$= 2\times 1 km$$d=2km$

Question 6

Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by which of the following options ?

Accepted Answer

Acceleration due to gravity,         $g = \dfrac{GMr}{R^3}$         for  $0\leq r\leq R$$\implies$    $g \propto  r$         for  $0\leq r\leq R$Acceleration due to gravity,         $g = \dfrac{GM}{r^2}$         for  $ r\geq R$$\implies$    $g \propto \dfrac{1}{r^2} $         for  $ r\geq R$Hence option C is correct.

Question 7

If ${v}_{e}$ is escape velocity and ${v}_{o}$ is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by :

Accepted Answer

${v}_{escape}=\sqrt {\cfrac{2GM}{R}}$${v}_{orbital}=\sqrt {\cfrac{GM}{R}}$${v}_{escape}=\sqrt {2}{v}_{o}$

Question 8

A particle of mass $m$ is thrown upwards from the surface of the earth, with a velocity $u$. The mass and the radius of the earth are, respectively, $M$ and $R$. $G$ is gravitational constant and $g$ is acceleration due to gravity on the surface of the earth. The minimum value of $u$ so that the particle does not return back to earth is

Accepted Answer

$GM = g{R}^{2}$${V}_{e} = \sqrt{2gR} = \sqrt{2\displaystyle\frac{GM}{{R}^{2}}}R = \sqrt{\displaystyle\frac{2GM}{R}}$.

Gravitation

Physics
13441 Views

A body weighs $72 N$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

A body of mass ' $m$ ' taken from the earth's surface to the height equal to twice the radius ( $R$ ) of the earth. The change in potential energy of body will be

A body weighs 200 N on the surface of the earth. How much it weigh half way down to the centre of the earth ?

The ratio of escape velocity at earth $(v_{e})$ to the escape velocity at a planet $(v_{p})$ whose radius and mean density are twice as that of earth is:

The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by which of the following options ?

If $v_{e}$ is escape velocity and $v_{o}$ is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by :

Gravitation

Physics 13441 Views

A body weighs 72N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

A body of mass 'm' taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be

A body weighs 200 N on the surface of the earth. How much it weigh half way down to the centre of the earth ?

The ratio of escape velocity at earth(ve​) to the escape velocity at a planet(vp​) whose radius and mean density are twice as that of earth is:

The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:

Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by which of the following options ?

If ve​ is escape velocity and vo​ is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by :

Physics
13441 Views

A body weighs $72 N$ on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

A body of mass ' $m$ ' taken from the earth's surface to the height equal to twice the radius ( $R$ ) of the earth. The change in potential energy of body will be

The ratio of escape velocity at earth $(v_{e})$ to the escape velocity at a planet $(v_{p})$ whose radius and mean density are twice as that of earth is:

If $v_{e}$ is escape velocity and $v_{o}$ is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by :