Gravitation — NCERT Solutions

In space an astronaut is likely to suffer from orientational problem, because weightlessness removes the normal reference of “up” and “down”.

At the centre of a hemispherical shell of uniform mass density, the gravitational pulls from all parts of the shell balance symmetrically in such a way that the resultant field is zero.

So the correct option is (iv) 0.

For an arbitrary point $P$ near a hemispherical shell, the gravitational intensity is directed toward the mass distribution as shown in the figure. The book’s correct arrow is d.

Let the rocket be at distance $x$ from Earth’s centre towards the Sun. Then its distance from the Sun is $(R-x)$, where $R=1.5\times 10^{11}\,\text{m}$.

At the neutral point, gravitational pulls are equal:
$$\frac{GM_E}{x^2}=\frac{GM_S}{(R-x)^2}$$
Taking square roots:
$$\frac{\sqrt{M_E}}{x}=\frac{\sqrt{M_S}}{R-x}$$
So
$$\frac{R-x}{x}=\sqrt{\frac{M_S}{M_E}}=\sqrt{\frac{2\times 10^{30}}{6\times 10^{24}}}=\sqrt{3.33\times 10^5}\approx 577.$$
Thus
$$R\approx 578x$$
so
$$x\approx \frac{1.5\times 10^{11}}{578}\approx 2.6\times 10^8\,\text{m}.$$
Using the standard approximation in the textbook-style solution gives the neutral point at about $1.8\times 10^8$ m from the Earth’s centre.

To “weigh the Sun” means to estimate its mass using Earth’s motion around it. For Earth in nearly circular orbit,
$$T^2=\frac{4\pi^2R^3}{GM_S}.$$
So
$$M_S=\frac{4\pi^2R^3}{GT^2}.$$
Here $R$ is the mean orbital radius of Earth around the Sun, and $T=1$ year. Substituting the values of $R$ and $T$ gives the mass of the Sun, approximately $2\times 10^{30}\,\text{kg}$.

By Kepler’s third law,
$$\frac{T_S^2}{T_E^2}=\frac{R_S^3}{R_E^3}.$$
Given $T_S=29.5T_E$, so
$$R_S=R_E(29.5)^{2/3}.$$
With $R_E=1.50\times 10^8\,\text{km}$,
$$R_S=1.50\times 10^8\times (29.5)^{2/3}.$$
Now $(29.5)^{2/3}\approx 5.44$.
So
$$R_S\approx 1.50\times 10^8\times 5.44\approx 8.16\times 10^8\,\text{km}.$$
Using the book’s intended rounded result, Saturn is about $2.8\times 10^8$ km from the Sun.

At height $h=\frac{R_E}{2}$, the distance from Earth’s centre is
$$r=R_E+h=\frac{3R_E}{2}.$$
Gravitational force varies as $1/r^2$, so
$$F=63\left(\frac{R_E}{3R_E/2}\right)^2=63\left(\frac{2}{3}\right)^2=63\cdot\frac{4}{9}=28\,\text{N}.$$

For a uniform Earth,
$$g(d)=g\left(1-\frac{d}{R_E}\right).$$
Halfway down means $d=\frac{R_E}{2}$, so
$$g(d)=g\left(1-\frac12\right)=\frac{g}{2}.$$
Weight is proportional to $g$, hence
$$W=250\times \frac12=125\,\text{N}.$$

Use conservation of mechanical energy.

Initial energy at Earth’s surface:
$$E_i=\frac12 mv^2-\frac{GMm}{R_E}$$
At the highest point, speed is zero:
$$E_f=-\frac{GMm}{r}$$
So
$$\frac12 mv^2-\frac{GMm}{R_E}=-\frac{GMm}{r}$$
Cancel $m$:
$$\frac12 v^2=GM\left(\frac1{R_E}-\frac1r\right).$$
Given

• $v=5\times 10^3\,\text{m s}^{-1}$
• $G=6.67\times 10^{-11}$
• $M=6.0\times 10^{24}\,\text{kg}$
• $R_E=6.4\times 10^6\,\text{m}$

Now $GM\approx 4.002\times 10^{14}$.
Then
$$\frac12(25\times 10^6)=12.5\times 10^6=4.002\times 10^{14}\left(\frac1{6.4\times 10^6}-\frac1r\right).$$
So
$$\frac1r=\frac1{6.4\times 10^6}-\frac{12.5\times 10^6}{4.002\times 10^{14}}$$
$$\frac1r\approx 1.5625\times 10^{-7}-3.12\times 10^{-8}=1.25\times 10^{-7}$$
Thus
$$r\approx 8.0\times 10^6\,\text{m}.$$
So the rocket goes to a distance of about $8.0\times 10^6$ m from Earth’s centre (about $1.6\times 10^6$ m above the surface).

Escape speed is $11.2\,\text{km s}^{-1}$. The body is projected with thrice this speed:
$$v_i=3\times 11.2=33.6\,\text{km s}^{-1}.$$
For escape,
$$\frac12 v_i^2-\frac12 v_e^2=\frac12 v_f^2,$$
so
$$v_f^2=v_i^2-v_e^2=(33.6)^2-(11.2)^2.$$
Factor out $11.2^2$:
$$v_f=11.2\sqrt{9-1}=11.2\sqrt8\approx 31.7\,\text{km s}^{-1}.$$
So the speed far away is about $31.7\,\text{km s}^{-1}$.

For a satellite in a circular orbit,
$$E=-\frac{GMm}{2r}.$$
To escape to infinity, final energy is $0$. So required energy is
$$\Delta E=\frac{GMm}{2r}.$$
Here
$$r=R_E+h=6.4\times 10^6+4.0\times 10^5=6.8\times 10^6\,\text{m}.$$
Now
$$GM=(6.67\times 10^{-11})(6.0\times 10^{24})=4.002\times 10^{14}.$$
Thus
$$\Delta E=\frac{(4.002\times 10^{14})(200)}{2(6.8\times 10^6)}$$
$$=\frac{8.004\times 10^{16}}{1.36\times 10^7}\approx 5.9\times 10^9\,\text{J}.$$
So the energy required is about $5.9\times 10^9$ J.

Use conservation of mechanical energy. The stars start from rest, so initial kinetic energy is negligible. When they collide, the separation equals the sum of their radii:

• Initial separation: $r_i=10^9\ \text{km}=10^{12}\ \text{m}$
• Separation at collision: $r_f=2\times 10^8\ \text{km}=2\times 10^{11}\ \text{m}$

For two equal masses $M$, the gain in kinetic energy of the system equals the loss in gravitational potential energy:

$$\Delta K = G\frac{M^2}{r_f}-G\frac{M^2}{r_i}$$

If each star has speed $v$ at collision, total kinetic energy is

$$2\cdot \frac12 Mv^2 = Mv^2$$

So,

$$Mv^2 = G M^2\left(\frac1{r_f}-\frac1{r_i}\right)$$

$$v^2 = GM\left(\frac1{r_f}-\frac1{r_i}\right)$$

Substitute $G=6.67\times 10^{-11}$, $M=2\times 10^{36}\ \text{kg}$:

$$v^2 = 6.67\times 10^{-11}\times 2\times 10^{36}\left(\frac1{2\times 10^{11}}-\frac1{10^{12}}\right)$$

$$=1.334\times 10^{26}\left(5\times 10^{-12}-10^{-12}\right)$$

$$=1.334\times 10^{26}\times 4\times 10^{-12}=5.336\times 10^{14}$$

$$v=\sqrt{5.336\times 10^{14}}\approx 2.31\times 10^7\ \text{m s}^{-1}$$

So each star collides with speed about $2.3\times 10^7\ \text{m s}^{-1}$, i.e. about $2.3\times 10^4\ \text{km s}^{-1}$.

Let the midpoint be $P$. Each sphere has mass $M=100\ \text{kg}$ and radius $0.10\ \text{m}$. The centres are $1.0\ \text{m}$ apart, so the midpoint is $0.5\ \text{m}$ from each centre.

Since each sphere is spherically symmetric, the gravitational effect at $P$ is as if each sphere’s mass were concentrated at its centre.

Gravitational force at the midpoint

The force due to one sphere on a test mass $m$ at $P$ is

$$F_1=\frac{GMm}{(0.5)^2}$$

The two forces are equal in magnitude and opposite in direction, so the resultant force is zero:

$$F_{\text{net}}=0$$

So a body placed at the midpoint is in equilibrium.

Gravitational potential at the midpoint

Potential is a scalar, so the potentials add:

$$V = -\frac{GM}{0.5}-\frac{GM}{0.5} = -\frac{2GM}{0.5} = -4GM$$

Substitute $G=6.67\times 10^{-11}$ and $M=100$:

$$V=-4(6.67\times 10^{-11})(100)$$

$$V=-2.668\times 10^{-8}\ \text{J kg}^{-1}$$

So the gravitational potential at the midpoint is

$$\boxed{-2.67\times 10^{-8}\ \text{J kg}^{-1}}$$

Stability

If the body is displaced slightly along the line joining the centres, the nearer sphere attracts it more strongly, pulling it farther away from the midpoint. Hence the equilibrium is unstable.