As the satellite moves to a higher orbit, its gravitational potential energy increases due to the increase in distance from the Earth's center.

The gravitational force decreases with the square of the distance. At 4R, the force is 1/(4^2) = 1/16 of the force at the surface.

Orbital speed v = √(GM/R). If R is tripled, v becomes √(GM/(3R)) = v/√3, which is reduced to one-third.

If the speed of a satellite is doubled, the radius of its orbit decreases by a factor of four due to the relationship between speed and radius in circular motion.

If the speed of a satellite is doubled, the orbital radius will decrease by a factor of four, as orbital speed is inversely proportional to the square root of the radius.

Increasing the speed of a satellite in a circular orbit will cause its orbit to become elliptical.

Orbital speed (v) = √(GM/r). As r increases, v decreases.

Gravitational force ∝ 1/r². If radius is doubled, force becomes 1/(2²) = 1/4.

Speed (v) = √(g*r) = √(9.8 m/s² * 7000 * 1000 m) = 7.9 km/s.

The gravitational force is inversely proportional to the square of the distance; halving the radius increases the force by a factor of four.

Centripetal acceleration a_c = v²/r = (7.9 km/s)² / (7000 km) = 0.00063 km/s² = 6.3 m/s².

Orbital speed v = √(GM/r). If r is tripled, v decreases by a factor of √3.

Increasing the speed of a satellite in a circular orbit will cause it to move into an elliptical orbit.

As the satellite moves to a higher orbit, its distance from the Earth increases, leading to an increase in gravitational potential energy.

Angular momentum L = mvr, where v is the orbital speed and r is the radius of the orbit.

The gravitational potential energy of a satellite in orbit is negative due to the attractive force of gravity.

The orbital speed of a satellite is given by v = sqrt(G * M / r), where M is the mass of the Earth.

The orbital speed of a satellite is given by v = sqrt(G * M / r), where M is the mass of the Earth.

The orbital speed can be calculated using the formula v = √(GM/r). For a height of 300 km, the speed is approximately 7.9 km/s.

A polar orbit allows the satellite to pass over the entire surface of the Earth as the planet rotates beneath it.

The gravitational acceleration experienced by a satellite at a height of 500 km is approximately 8.7 m/s².

The time period (T) is the reciprocal of frequency (f). T = 1/f = 1/5 = 0.2 s.

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 200 N/m and m = 0.5 kg. Thus, ω = √(200/0.5) = √400 = 20 rad/s.

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 200 N/m and m = 2 kg. Thus, ω = √(200/2) = √100 = 10 rad/s.

The angular frequency ω is given by the formula ω = √(k/m). Here, k = 50 N/m and m = 2 kg. Thus, ω = √(50/2) = √25 = 5 rad/s.

ω = V_max/A = 2/0.1 = 20 rad/s.

Maximum potential energy (PE) = (1/2)kA^2 = (1/2)(200)(0.1^2) = 1 J.

T = 2π√(m/k) = 2π√(2/200) = 1 s.

The total energy E is proportional to the square of the amplitude. If the amplitude is halved, the new energy will be E/4.

Total energy E = (1/2)kA^2. 50 = (1/2)k(0.1^2) => k = 500 N/m.