A particle is projected from pointA, that is at a distance4Rfrom the centre of the earth, with speedV1in a direction making 30° with the line joining the centre of the earth and pointA, as shown in the figure. Find the speedV1.if particle passes grazing the surface of the earth. Consider gravitational interaction only between these two. (useGM/R= 6.4 ×107m2/s2)Express your answer in the form of (X/√2)km/sec and fill value ofX.Correct answer is '8'. Can you explain this answer?

Question

A particle is projected from pointA, that is at a distance4Rfrom the centre of the earth, with speedV1in a direction making 30&deg; with the line joining the centre of the earth and pointA, as shown in the figure. Find the speedV1.if particle passes grazing the surface of the earth. Consider gravitational interaction only between these two. (useGM/R= 6.4 &times;107m2/s2)Express your answer in the form of (X/&radic;2)km/sec and fill value ofX.Correct answer is '8'. Can you explain this answer?

Answer

Degrees with the horizontal. The particle then moves to a point B at a distance R from the center of the earth. Find the speed of the particle at point B.

To solve this problem, we can use the principle of conservation of mechanical energy. At point A, the particle has gravitational potential energy and kinetic energy. At point B, the particle has gravitational potential energy and kinetic energy as well. Since the particle is not acted upon by any external forces, the total mechanical energy is conserved.

At point A, the gravitational potential energy is given by mgh, where m is the mass of the particle, g is the acceleration due to gravity, and h is the height of the particle above the surface of the earth. We can calculate h using the Pythagorean theorem:

h = √((4R)^2 - R^2) = √(16R^2 - R^2) = √(15R^2) = √15R

The kinetic energy at point A is given by (1/2)mv1^2, where v1 is the initial speed of the particle.

At point B, the gravitational potential energy is given by mgh, where h is the height of the particle above the surface of the earth. Since the particle is at a distance R from the center of the earth, h = R.

The kinetic energy at point B is given by (1/2)mv2^2, where v2 is the speed of the particle at point B.

Since the total mechanical energy is conserved, we have:

mgh + (1/2)mv1^2 = mgh + (1/2)mv2^2

Canceling out the mass (m) on both sides of the equation, we get:

gh + (1/2)v1^2 = gh + (1/2)v2^2

Simplifying the equation, we have:

(1/2)v1^2 = (1/2)v2^2

Taking the square root of both sides, we get:

v1 = v2

Therefore, the speed of the particle at point B is equal to the initial speed v1.

Answer

let v₂ to be speed when it grazes earth

Conserving angular momentum we get
mv₁sin30^o4R = m.v₂R.....(i)
Conserving total energy we get
1/2mV₁² − GmM/4R

= 1/2 mV₂²− GmM/R

putting V₂= V₁sin30^o× 4
= 2V₁ we get
1/2mV₁²(1−4) = −3GmM/4R
or

V₁ =(√Gm/2R)

= (√6.4*10⁷)/2

= 500X√2

X = (8000)/(500*2)

X = 8

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