Solution:

The escape velocity \( V_e \) from the Earth's surface is given by:

\[
V_e = \sqrt{\frac{2GM}{R}}
\]

At a height \( h = R \) (i.e., the platform is at a distance \( 2R \) from the Earth's center), the escape velocity \( V \) becomes:

\[
V = \sqrt{\frac{2GM}{2R}} = \frac{V_e}{\sqrt{2}}
\]

So, the factor \( F \) is:

\[
F = \frac{V}{V_e} = \frac{1}{\sqrt{2}}
\]

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Solution:

Gravitational potential \( V \) at the midpoint is the sum of the potentials due to both masses:

\[
V = -\frac{G \cdot 10^2}{0.5} - \frac{G \cdot 10^3}{0.5}
\]

Simplifying:

\[
V = -2G(10^2 + 10^3) = -2G \cdot 1100
\]

So, the gravitational potential at the midpoint is:

\[
V = -2200G
\]

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Solution:

The gravitational potential inside a spherical shell (including at the center) due to the shell is constant. The potential at a distance \( a/2 \) from the center is the sum of the potentials due to the mass at the center and the shell.

\[
V = -\frac{GM}{a/2} - \frac{GM}{a}
\]

Simplifying:

\[
V = -\frac{2GM}{a} - \frac{GM}{a} = -\frac{3GM}{a}
\]

So, the potential at \( a/2 \) is:

\[
V = -\frac{3GM}{a}
\]

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Solution:

The gravitational potential \( V \) at height \( h \) from the Earth's surface is given by:

\[
V = -\frac{GM}{R + h}
\]

The acceleration due to gravity \( g \) at height \( h \) is:

\[
g = \frac{GM}{(R + h)^2}
\]

Given:
- \( V = -5.4 \times 10^7 \, \text{J/kg} \)
- \( g = 6.0 \, \text{m/s}^2 \)
- \( R = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \)

From the first equation:

\[
-5.4 \times 10^7 = -\frac{GM}{R + h}
\]

From the second equation:

\[
6.0 = \frac{GM}{(R + h)^2}
\]

Dividing the two equations to eliminate \( GM \):

\[
\frac{V}{g} = \frac{-(R + h)}{(R + h)^2}
\]

Simplifying:

\[
\frac{-5.4 \times 10^7}{6.0} = -(R + h)
\]

Solving for \( h \):

\[
R + h = 9 \times 10^6 \, \text{m}
\]

\[
h = 9 \times 10^6 - 6.4 \times 10^6 = 2.6 \times 10^6 \, \text{m} = 2600 \, \text{km}
\]

Thus, the height is \( 2600 \, \text{km} \).

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Solution:

The velocity required to reach a height equal to the Earth's radius \( R \) is the escape velocity for a total distance of \( 2R \) from the Earth's center.

Escape velocity formula:

\[
v = \sqrt{\frac{GM}{R}}
\]

At height \( h = R \), the velocity needed is:

\[
v = \sqrt{\frac{GM}{2R}} = \frac{V_e}{\sqrt{2}}
\]

Thus, the velocity is:

\[
v = \frac{V_e}{\sqrt{2}} = \frac{\sqrt{2GM/R}}{\sqrt{2}} = \sqrt{\frac{GM}{R}} = V_e/\sqrt{2}= \sqrt{\frac{GM}{R}}
\]

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Solution:

The point where the gravitational field is zero lies closer to the smaller mass \( m \). Let the distance of this point from \( m \) be \( x \), and from \( 4m \) be \( r - x \).

At this point:

\[
\frac{Gm}{x^2} = \frac{G \cdot 4m}{(r - x)^2}
\]

Taking the square root:

\[
\frac{1}{x} = \frac{2}{r - x}
\]

Solving:

\[
r - x = 2x \quad \Rightarrow \quad x = \frac{r}{3}
\]

The gravitational potential \( V \) at this point is the sum of the potentials due to both masses:

\[
V = -\frac{Gm}{\frac{r}{3}} - \frac{G \cdot 4m}{\frac{2r}{3}} = -\frac{3Gm}{r} - \frac{6Gm}{r} = -\frac{9Gm}{r}
\]

Thus, the potential at the point is:

\[
V = -\frac{9Gm}{r}
\]

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Solution:

The gravitational potential at the center due to a semicircular rod is given by:

\[
V = - \frac{GM}{R}
\]

Where \( R \) is the radius of the semicircle, which is related to the length of the rod:

\[
L = \pi R \quad \Rightarrow \quad R = \frac{L}{\pi}
\]

Substituting \( R \) into the potential formula:

\[
V = - \frac{GM}{\frac{L}{\pi}} = - \frac{\pi GM}{L}
\]

Thus, the gravitational potential at the center is:

\[
V = - \frac{\pi GM}{L}
\]

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Solution:

The escape velocity \( V_e \) is the speed needed to escape Earth's gravitational pull. The energy conservation principle applies, where the total mechanical energy at the surface and at height \( h = 3R \) (where \( R \) is the Earth's radius) should be equal.

The total energy at the surface:
\[
E_1 = \frac{1}{2} m V_e^2 - \frac{G M m}{R}
\]

At height \( 3R \), the total energy is:
\[
E_2 = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Since \( E_1 = E_2 \), we equate the two:
\[
\frac{1}{2} m V_e^2 - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

We know that the escape velocity is given by:
\[
V_e^2 = \frac{2 G M}{R}
\]

Substitute this into the equation:
\[
\frac{1}{2} m \frac{2 G M}{R} - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Simplifying:
\[
\frac{G M m}{R} - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

This simplifies to:
\[
0 = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Solving for \( v^2 \):
\[
\frac{1}{2} m v^2 = \frac{G M m}{4R}
\]

\[
v^2 = \frac{2 G M}{4R} = \frac{G M}{2R}
\]

Thus, the velocity at height \( 3R \) is:
\[
v = \sqrt{\frac{G M}{2R}} = \frac{V_e}{\sqrt{2}}= 7.92 km/sec
\]

So, the velocity at height \( 3R \) is \( \frac{V_e}{\sqrt{2}} \).= 7.92 km/sec

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Solution:

Total gravitational potential at a point \( r = \frac{a}{2} \):

1. Potential due to the mass at the center:
The gravitational potential at a distance \( r \) from a point mass \( M \) is given by:
\[
V_{\text{center}} = -\frac{GM}{r}
\]
At \( r = \frac{a}{2} \), this becomes:
\[
V_{\text{center}} = -\frac{GM}{\frac{a}{2}} = -\frac{2GM}{a}
\]

2. **Potential due to the spherical shell**:
Inside a spherical shell, the gravitational potential is constant and equal to the potential at the surface, which is:
\[
V_{\text{shell}} = -\frac{GM}{a}
\]

### Total potential at \( r = \frac{a}{2} \):

The total gravitational potential is the sum of the potentials due to the mass at the center and the shell:
\[
V_{\text{total}} = V_{\text{center}} + V_{\text{shell}} = -\frac{2GM}{a} - \frac{GM}{a} = -\frac{3GM}{a}
\]

So, the gravitational potential at a distance \( \frac{a}{2} \) from the center is:
\[
V = -\frac{3GM}{a}
\]

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